Rs Aggarwal 2017 for Class 9 Math Chapter 1 - Real Numbers

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Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 238:

Question 1:

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Answer:

We have:
Base = 24 cm
Height = 14.5 cm

Now,
$Area of triangle = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 24 \times 14.5 = 174 {cm}^{2}$

Page No 238:

Question 2:

The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

Answer:

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle = $\frac{Total Cost}{Rate} = \frac{783}{58} = 13.5 ha = 135000 m^{2}$
We have:
$Area of triangle = 135000 m^{2} \Rightarrow \frac{1}{2} \times Base \times Heigh t = 135000 \Rightarrow \frac{1}{2} \times 3 h \times h = 135000 \Rightarrow h^{2} = \frac{135000 \times 2}{3} \Rightarrow h^{2} = 90000 \Rightarrow h = 300 m$

Thus, we have:
Height = h = 300 m
Base = 3h = 900 m

Page No 238:

Question 3:

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.

Answer:

$Let : a = 42 cm, b = 34 cm and c = 20 cm ∴ s = \frac{a + b + c}{2} = \frac{42 + 34 + 20}{2} = 48 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{48 (48 - 42) (48 - 34) (48 - 20)} = \sqrt{48 \times 6 \times 14 \times 28} = \sqrt{4 \times 2 \times 6 \times 6 \times 7 \times 2 \times 7 \times 4} = 4 \times 2 \times 6 \times 7 = 336 {cm}^{2}$

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:
$Area of triangle = 336 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 336 \Rightarrow Height = \frac{336 \times 2}{42} = 16 cm$

Page No 238:

Question 4:

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.

Answer:

$Let : a = 18 cm, b = 24 cm and c = 30 cm ∴ s = \frac{a + b + c}{2} = \frac{18 + 24 + 30}{2} = 36 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{36 (36 - 18) (36 - 24) (36 - 30)} = \sqrt{36 \times 18 \times 12 \times 6} = \sqrt{12 \times 3 \times 6 \times 3 \times 12 \times 6} = 12 \times 3 \times 6 = 216 {cm}^{2}$

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:
$Area of triangle = 216 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 216 \Rightarrow Height = \frac{216 \times 2}{18} = 24 cm$

Page No 238:

Question 5:

Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side.

Answer:

$Let : a = 91 m, b = 98 m and c = 105 m ∴ s = \frac{a + b + c}{2} = \frac{91 + 98 + 105}{2} = 147 m By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{147 (147 - 91) (147 - 98) (147 - 105)} = \sqrt{147 \times 56 \times 49 \times 42} = \sqrt{7 \times 3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \times 7 \times 3 \times 2} = 7 \times 7 \times 7 \times 2 \times 3 \times 2 = 4116 m^{2}$

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.
$Area of triangle = 4116 m^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 4116 \Rightarrow Height = \frac{4116 \times 2}{105} = 78.4 m$

Page No 238:

Question 6:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle.

Answer:

Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5 $\times$ 5 = 25 m
12 $\times$ 5 = 60 m
13 $\times$ 5 = 65 m

Now,
$Let : a = 25 m, b = 60 m and c = 65 m ∴ s = \frac{150}{2} = 75 m By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{75 (75 - 25) (75 - 60) (75 - 65)} = \sqrt{75 \times 50 \times 15 \times 10} = \sqrt{15 \times 5 \times 5 \times 10 \times 15 \times 10} = 15 \times 5 \times 10 = 750 m^{2}$

Page No 238:

Question 7:

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs 18.80 per 10 m².

Answer:

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25 $\times$ 10 = 250 m
17 $\times$ 10 = 170 m
12 $\times$ 10 = 120 m

Now,
$Let : a = 250 m, b = 170 m and c = 120 m ∴ s = \frac{540}{2} = 270 m By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{270 (270 - 250) (270 - 170) (270 - 120)} = \sqrt{270 \times 20 \times 100 \times 150} = \sqrt{30 \times 3 \times 3 \times 20 \times 20 \times 5 \times 30 \times 5} = 30 \times 3 \times 20 \times 5 = 9000 m^{2}$

Cost of ploughing 10 m² field = Rs 18.80
Cost of ploughing 1 m² field = Rs $\frac{18.80}{10}$

Cost of ploughing 9000 m² field = $\frac{18.80}{10} \times 9000 = Rs 16920$

Page No 238:

Question 8:

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

Answer:

$(i) Let : a = 85 m and b = 154 m Given : Perimeter = 324 m o r, a + b + c = 324 \Rightarrow c = 324 - 85 - 154 = 85 m ∴ s = \frac{324}{2} = 162 m By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{162 (162 - 85) (162 - 154) (162 - 85)} = \sqrt{162 \times 77 \times 8 \times 77} = \sqrt{1296 \times 77 \times 77} = \sqrt{36 \times 77 \times 77 \times 36} = 36 \times 77 = 2772 m^{2}$

(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:
$Area of triangle = 2772 m^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 2772 \Rightarrow Height = \frac{2772 \times 2}{154} = 36 m$

Page No 238:

Question 9:

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.

Answer:

$We have : a = 13 cm and b = 20 cm ∴ Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} = \frac{20}{4} \times \sqrt{4 (13)^{2} - 20^{2}} = 5 \times \sqrt{676 - 400} = 5 \times \sqrt{276} = 5 \times 16.6 = 83.06 {cm}^{2}$

Page No 238:

Question 10:

The base of an isosceles triangle measures 80 cm and its area is 360 cm². Find the perimeter of the triangle.

Answer:

Let $△$ PQR be an isosceles triangle and PX $⊥$ QR.
Now,
$Area of triangle = 360 {cm}^{2} \Rightarrow \frac{1}{2} \times Q R \times P X = 360 \Rightarrow h = \frac{720}{80} = 9 cm Now, Q X = \frac{1}{2} \times 80 = 40 cm and P X = 9 cm$
Also,
$P Q = \sqrt{Q X^{2} + P X^{2}} a = \sqrt{40^{2} + 9^{2}} = \sqrt{1600 + 81} = \sqrt{1681} = 41 cm$

∴ Perimeter = 80 + 41 + 41 = 162 cm

Page No 238:

Question 11:

The perimeter of an isosceles triangle is 42 cm and its base is $1 \frac{1}{2}$ times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle.

Answer:

Let the equal sides of the isosceles triangle be a cm each.
∴ Base of the triangle, b = $\frac{3}{2}$ a cm
(i) Perimeter = 42 cm
or, a + a + $\frac{3}{2}$ a = 42
or, 2a + $\frac{3}{2}$ a= 42

$\Rightarrow 2 a + \frac{3}{2} a = 42 \Rightarrow \frac{7 a}{2} = 42 \Rightarrow a = 12$

So, equal sides of the triangle are 12 cm each.
Also,
Base = $\frac{3}{2}$ a = $\frac{3}{2} \times 12 = 18 cm$
(ii)
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} = \frac{18}{4} \times \sqrt{4 (12)^{2} - 18^{2}} (a = 12 cm and b = 18 cm) = 4.5 \times \sqrt{576 - 324} = 4.5 \times \sqrt{252} = 4.5 \times 15.87 = 71.42 {cm}^{2}$

(iii)
$Area of triangle = 71.42 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 71.42 \Rightarrow Height = \frac{71.42 \times 2}{18} = 7.94 cm$

Page No 238:

Question 12:

If the area of an equilateral triangle is $36 \sqrt{3} {cm}^{2}$ , find its perimeter.

Answer:

$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} \Rightarrow \frac{\sqrt{3}}{4} \times (Side)^{2} = 36 \sqrt{3}$
$\Rightarrow (Side)^{2} = 144 \Rightarrow Side = 12 cm$

Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm

Page No 238:

Question 13:

If the area of an equilateral triangle is $81 \sqrt{3} {cm}^{2}$ , find its height.

Answer:

$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} \Rightarrow \frac{\sqrt{3}}{4} \times (Side)^{2} = 81 \sqrt{3}$
$\Rightarrow (Side)^{2} = 324 \Rightarrow Side = 18 cm$

Now, we have:
$Height = \frac{\sqrt{3}}{2} \times Side = \frac{\sqrt{3}}{2} \times 18 = 9 \sqrt{3} cm$

Page No 238:

Question 14:

The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle.

Answer:

Let $△$ PQR be a right-angled triangle and PQ $⊥$ QR.
Now,
$P Q = \sqrt{P R^{2} - Q R^{2}} = \sqrt{50^{2} - 48^{2}} = \sqrt{2500 - 2304} = \sqrt{196} = 14 cm$

$Area of triangle = \frac{1}{2} \times Q R \times P Q = \frac{1}{2} \times 48 \times 14 = 336 {cm}^{2}$

Page No 238:

Question 15:

Each side of an equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take $\sqrt{3} = 1.732$ .

Answer:

Side of the equilateral triangle = 8 cm

(i)
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (8)^{2} = \frac{1.732 \times 64}{4} = 27.71 {cm}^{2}$

(ii)
$Height = \frac{\sqrt{3}}{2} \times Side = \frac{\sqrt{3}}{2} \times 8 = \frac{1.732 \times 8}{2} = 6.93 cm$

Page No 238:

Question 16:

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take $\sqrt{3} = 1.732$ .

Answer:

Height of the equilateral triangle = 9 cm
Thus, we have:
$Height = \frac{\sqrt{3}}{2} \times Side \Rightarrow 9 = \frac{\sqrt{3}}{2} \times Side \Rightarrow Side = \frac{18}{\sqrt{3}} = \frac{18}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = 6 \sqrt{3} cm$

Also,
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (6 \sqrt{3})^{2} = \frac{108}{4} \sqrt{3} = 27 \sqrt{3} = 46.76 {cm}^{2}$

Page No 239:

Question 17:

An umbrella is made by stitching 12 triangular pieces of cloth, each measuring (50 cm × 20 cm × 50 cm). Find the area of the cloth used in it.

Answer:

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} = \frac{20}{4} \times \sqrt{4 (50)^{2} - 20^{2}} (a = 50 cm and b = 20 cm) = 5 \times \sqrt{10000 - 400} = 5 \times \sqrt{9600} = 5 \times 40 \sqrt{6} = 200 \sqrt{6} = 490 {cm}^{2}$

Now,
Area of 1 triangular piece of cloth = 490 cm²
Area of 12 triangular pieces of cloth = $12 \times 490 = 5880 {cm}^{2}$

Page No 239:

Question 18:

A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm.

Answer:

Area of one triangular-shaped tile can be found in the following manner:

$Let : a = 16 cm, b = 12 cm and c = 20 cm s = \frac{a + b + c}{2} = \frac{16 + 12 + 20}{2} = 24 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{24 (24 - 16) (24 - 12) (24 - 20)} = \sqrt{24 \times 8 \times 12 \times 4} = \sqrt{6 \times 4 \times 4 \times 4 \times 4 \times 6} = 6 \times 4 \times 4 = 96 {cm}^{2}$

Now,
Area of 16 triangular-shaped tiles = $16 \times 96 = 1536 {cm}^{2}$
Cost of polishing tiles of area 1 cm² = Rs 1
Cost of polishing tiles of area 1536 cm² = $1 \times 1536 = Rs 1536$

Page No 239:

Question 19:

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ∠ACB = 90° and AC = 15 cm.

Answer:

We know that $△$ ABC is a right-angled triangle.

∴ $B C^{2} = \sqrt{A B^{2} - A C^{2}} = \sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225} = \sqrt{64} = 8 cm$

$Now, Area of triangle A B C = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times B C \times A C = \frac{1}{2} \times 8 \times 15 = 60 {cm}^{2}$

$Let : a = 9 cm, b = 15 cm and c = 12 cm s = \frac{a + b + c}{2} = \frac{9 + 15 + 12}{2} = 18 cm By Heron' s formula, we have : Area of triangle A C D = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{18 (18 - 9) (18 - 15) (18 - 12)} = \sqrt{18 \times 9 \times 3 \times 6} = \sqrt{6 \times 3 \times 3 \times 3 \times 3 \times 6} = 6 \times 3 \times 3 = 54 {cm}^{2}$

Now,
Area of quadrilateral ABCD = Area of $△$ ABC + Area of $△$ ACD
= (60 + 54) cm² =114 cm²
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm

Page No 239:

Question 20:

Find the perimeter and area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and ∠CBD = 90°.

Answer:

We know that $△$ DBC is a right-angled triangle.
$∴ Area of triangle D B C = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times B C \times B D = \frac{1}{2} \times 20 \times 21 = 210 {cm}^{2}$

Now,
$Let : a = 42 cm, b = 34 cm and c = 20 cm s = \frac{a + b + c}{2} = \frac{42 + 34 + 20}{2} = 48 cm By Heron' s formula, we have : Area of triangle A B D = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{48 (48 - 42) (48 - 34) (48 - 20)} = \sqrt{48 \times 6 \times 14 \times 28} = \sqrt{4 \times 2 \times 6 \times 6 \times 7 \times 2 \times 7 \times 4} = 4 \times 2 \times 6 \times 7 = 336 {cm}^{2}$

∴ Area of quadrilateral ABCD = Area of $△$ DBC + Area of $△$ ABD
=(210 + 336) cm² = 546 cm²
Also,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 42 + 21 + 29 + 34 = 126 cm

Page No 239:

Question 21:

Find the area of quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and ∆BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given: $\sqrt{3} = 1.73]$ ]

Answer:

We know that $△$ BAD is a right-angled triangle.

∴ $A B = \sqrt{B D^{2} - A D^{2}} = \sqrt{26^{2} - 24^{2}} = \sqrt{676 - 576} = \sqrt{100} = 10 cm$

$Now, Area of triangle B A D = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times A B \times A D = \frac{1}{2} \times 10 \times 24 = 120 {cm}^{2}$

Also, we know that $△$ BDC is an equilateral triangle.
$∴ Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (26)^{2} = \frac{\sqrt{3}}{4} \times 676 = 169 \sqrt{3} = 292.37 {cm}^{2}$

Now,
Area of quadrilateral ABCD = Area of $△$ ABD + Area of $△$ BDC
= (120 + 292.37) cm² = 412.37 cm²

Page No 239:

Question 22:

Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm.

Answer:

$Let : a = 26 cm, b = 30 cm and c = 28 cm s = \frac{a + b + c}{2} = \frac{26 + 30 + 28}{2} = 42 cm By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{42 (42 - 26) (42 - 30) (42 - 28)} = \sqrt{42 \times 16 \times 12 \times 14} = \sqrt{14 \times 3 \times 4 \times 4 \times 2 \times 2 \times 3 \times 14} = 14 \times 4 \times 2 \times 3 = 336 {cm}^{2}$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2 \times 336 = 672 {cm}^{2}$

Page No 240:

Question 23:

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: $\sqrt{3} = 1.73$ ]

Answer:

$Let : a = 10 cm, b = 16 cm and c = 14 cm s = \frac{a + b + c}{2} = \frac{10 + 16 + 14}{2} = 20 cm By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{20 (20 - 10) (20 - 16) (20 - 14)} = \sqrt{20 \times 10 \times 4 \times 6} = \sqrt{10 \times 2 \times 10 \times 2 \times 2 \times 3 \times 2} = 10 \times 2 \times 2 \sqrt{3} = 69.2 {cm}^{2}$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2 \times 69.2 {cm}^{2} = 138.4 {cm}^{2}$

Page No 240:

Question 24:

In the given figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL ⊥ BD and CM ⊥ BD such that AL = 16.8 cm and CM = 13.2 cm. Calculate the area of quadrilateral ABCD.

Answer:

$Area of A B C D = Area of △ A B D + Area of △ B D C = \frac{1}{2} \times B D \times A L + \frac{1}{2} \times B D \times C M = \frac{1}{2} \times B D (A L + C M) = \frac{1}{2} \times 64 (16.8 + 13.2) = 32 \times 30 = 960 {cm}^{2}$

Page No 240:

Question 1:

In a ∆ABC it is given that base = 12 cm and height = 5 cm. Its area is
(a) 60 cm²
(b) 30 cm²
(c) $15 \sqrt{3} {cm}^{2}$
(d) 45 cm²

Answer:

(b) 30 cm²

$Area of triangle = \frac{1}{2} \times Base \times Height Area of ∆ A B C = \frac{1}{2} \times 12 \times 5 = 30 {cm}^{2}$

Page No 240:

Question 2:

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
(a) 96 cm²
(b) 120 cm²
(c) 144 cm²
(d) 160 cm²

Answer:

(a) 96 cm²

$Let : a = 20 cm, b = 16 cm and c = 12 cm s = \frac{a + b + c}{2} = \frac{20 + 16 + 12}{2} = 24 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{24 (24 - 20) (24 - 16) (24 - 12)} = \sqrt{24 \times 4 \times 8 \times 12} = \sqrt{6 \times 4 \times 4 \times 4 \times 4 \times 6} = 6 \times 4 \times 4 = 96 {cm}^{2}$

Page No 240:

Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the triangle is
(a) $8 \sqrt{3} {cm}^{2}$
(b) $16 \sqrt{3} {cm}^{2}$
(c) $32 \sqrt{3} {cm}^{2}$
(d) 48 cm²

Answer:

(b) $16 \sqrt{3} {cm}^{2}$
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (8)^{2} = \frac{\sqrt{3}}{4} \times 64 = 16 \sqrt{3} {cm}^{2}$

Page No 240:

Question 4:

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
(a) $16 \sqrt{5} {cm}^{2}$
(b) $8 \sqrt{5} {cm}^{2}$
(c) $16 \sqrt{3} {cm}^{2}$
(d) $8 \sqrt{3} {cm}^{2}$

Answer:

(b) $8 \sqrt{5} {cm}^{2}$
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 6 cm and b = 8 cm Thus, we have : \frac{8}{4} \times \sqrt{4 (6)^{2} - 8^{2}} = \frac{8}{4} \times \sqrt{144 - 64} = \frac{8}{4} \times \sqrt{80} = \frac{8}{4} \times 4 \sqrt{5} = 8 \sqrt{5} {cm}^{2}$

Page No 241:

Question 5:

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
(a) 8 cm
(b) $\sqrt{30} cm$
(c) 4 cm
(d) $\sqrt{11} cm$

Answer:

(c) 4 cm
$Height of isosceles triangle = \frac{1}{2} \sqrt{4 a^{2} - b^{2}} = \frac{1}{2} \sqrt{4 {(5)}^{2} - 6^{2}} (a = 5 cm and b = 6 cm) = \frac{1}{2} \times \sqrt{100 - 36} = \frac{1}{2} \times \sqrt{64} = \frac{1}{2} \times 8 = 4 cm$

Page No 241:

Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a) $5 \sqrt{10} {cm}^{2}$
(b) 50 cm²
(c) $10 \sqrt{3} {cm}^{2}$
(d) 75 cm²

Answer:

(b) 50 cm²
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
$Area of triangle = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 10 \times 10 = 50 {cm}^{2}$

Page No 241:

Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the triangle is
(a) $10 \sqrt{3} cm$
(b) $5 \sqrt{3} cm$
(c) $10 \sqrt{2} cm$
(d) 5 cm

Answer:

(b) $5 \sqrt{3} cm$
$Height of equilateral triangle = \frac{\sqrt{3}}{2} \times Side = \frac{\sqrt{3}}{2} \times 10 = 5 \sqrt{3} cm$

Page No 241:

Question 8:

The height of an equilateral triangle is 6 cm. Its area is
(a) $12 \sqrt{3} {cm}^{2}$
(b) $6 \sqrt{3} {cm}^{2}$
(c) $12 \sqrt{2} {cm}^{2}$
(d) 18 cm²

Answer:

(a) $12 \sqrt{3} {cm}^{2}$
$Height of equilateral triangle = \frac{\sqrt{3}}{2} \times Side \Rightarrow 6 = \frac{\sqrt{3}}{2} \times Side \Rightarrow Side = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12}{3} \times \sqrt{3} = 4 \sqrt{3} cm Now, Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times {(4 \sqrt{3})}^{2} = \frac{\sqrt{3}}{4} \times 48 = 12 \sqrt{3} {cm}^{2}$

Page No 241:

Question 9:

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
(a) 480 m²
(b) 320 m²
(c) 384 m²
(d) 360 m²

Answer:

(c) 384 m²

$Let : a = 40 m, b = 24 m and c = 32 m s = \frac{a + b + c}{2} = \frac{40 + 24 + 32}{2} = 48 m By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{48 (48 - 40) (48 - 24) (48 - 32)} = \sqrt{48 \times 8 \times 24 \times 16} = \sqrt{24 \times 2 \times 8 \times 24 \times 8 \times 2} = 24 \times 8 \times 2 = 384 m^{2}$

Page No 241:

Question 10:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
(a) 375 cm²
(b) 750 cm²
(c) 250 cm²
(d) 500 cm²

Answer:

(b) 750 cm²

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5 $\times$ 5 cm, 12 $\times$ 5 cm and 13 $\times$ 5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,
$Let : a = 25 cm, b = 60 cm and c = 65 cm s = \frac{150}{2} = 75 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{75 (75 - 25) (75 - 60) (75 - 65)} = \sqrt{75 \times 50 \times 15 \times 10} = \sqrt{15 \times 5 \times 5 \times 10 \times 15 \times 10} = 15 \times 5 \times 10 = 750 {cm}^{2}$

Page No 241:

Question 11:

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
(a) 24 cm
(b) 18 cm
(c) 30 cm
(d) 12 cm

Answer:

(a) 24 cm

$Let : a = 30 cm, b = 24 cm and c = 18 cm s = \frac{a + b + c}{2} = \frac{30 + 24 + 18}{2} = 36 cm On applying Heron' s formula, we get : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{36 (36 - 30) (36 - 24) (36 - 18)} = \sqrt{36 \times 6 \times 12 \times 18} = \sqrt{12 \times 3 \times 12 \times 6 \times 3} = 12 \times 3 \times 6 = 216 {cm}^{2}$

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:
$Area of triangle = 216 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 216 \Rightarrow Height = \frac{216 \times 2}{18} = 24 cm$

Page No 241:

Question 12:

The base of an isosceles triangle is 16 cm and its area is 48 cm². The perimeter of the triangle is
(a) 41 cm
(b) 36 cm
(c) 48 cm
(d) 324 cm

Answer:

(b) 36 cm

Let $△$ PQR be an isosceles triangle and PX $⊥$ QR.
Now,
$Area of triangle = 48 {cm}^{2} \Rightarrow \frac{1}{2} \times Q R \times P X = 48 \Rightarrow h = \frac{96}{16} = 6 cm Also, Q X = \frac{1}{2} \times 24 = 12 cm and P X = 12 cm$
$P Q = \sqrt{Q X^{2} + P X^{2}} a = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 cm$

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

Page No 241:

Question 13:

The area of an equilateral triangle is $36 \sqrt{3} {cm}^{2}$ . Its perimeter is
(a) 36 cm
(b) $12 \sqrt{3} cm$
(c) 24 cm
(d) 30 cm

Answer:

(a) 36 cm
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} \Rightarrow \frac{\sqrt{3}}{4} \times (Side)^{2} = 36 \sqrt{3}$

$\Rightarrow (Side)^{2} = 144 \Rightarrow Side = 12 cm$

Now,
Perimeter = 3 × Side = 3 × 12 = 36 cm

Page No 241:

Question 14:

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 156 cm²
(b) 78 cm²
(c) 60 cm²
(d) 120 cm²

Answer:

(c) 60 cm²
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 13 cm and b = 24 cm Thus, we have : \frac{24}{4} \times \sqrt{4 (13)^{2} - 24^{2}} = 6 \times \sqrt{676 - 576} = 6 \times \sqrt{100} = 6 \times 10 = 60 {cm}^{2}$

Page No 241:

Question 15:

The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is
(a) 168 cm²
(b) 252 cm²
(c) 336 cm²
(d) 504 cm²

Answer:

(c) 336 cm²

Let $△$ PQR be a right-angled triangle and PQ $⊥$ QR.
Now,
$P Q = \sqrt{P R^{2} - Q R^{2}} = \sqrt{50^{2} - 48^{2}} = \sqrt{2500 - 2304} = \sqrt{196} = 14 cm$

$∴ Area of triangle = \frac{1}{2} \times Q R \times P Q = \frac{1}{2} \times 48 \times 14 = 336 {cm}^{2}$

Page No 241:

Question 16:

The area of an equilateral triangle is $81 \sqrt{3} {cm}^{2}$ . Its height is
(a) $9 \sqrt{3} cm$
(b) $6 \sqrt{3} cm$
(c) $18 \sqrt{3} cm$
(d) 9 cm

Answer:

(a) $9 \sqrt{3} cm$
$Area of equilateral triangle = 81 \sqrt{3} {cm}^{2} \Rightarrow \frac{\sqrt{3}}{4} \times (Side)^{2} = 81 \sqrt{3} \Rightarrow (Side)^{2} = 81 \times 4 \Rightarrow (Side)^{2} = 324 \Rightarrow Side = 18 cm Now, Height = \frac{\sqrt{3}}{2} \times Side = \frac{\sqrt{3}}{2} \times 18 = 9 \sqrt{3} cm$

Page No 242:

Question 17:

The difference between the semi-perimeter and the sides of a ∆ABC are 8 cm, 7 cm and 5 cm respectively. The area of the triangle is
(a) $20 \sqrt{7} {cm}^{2}$
(b) $10 \sqrt{14} {cm}^{2}$
(c) $20 \sqrt{14} {cm}^{2}$
(d) 140 cm²

Answer:

(c) $20 \sqrt{14} {cm}^{2}$
Let the sides of the triangle be a cm, b cm and c cm and the semi-perimeter be s cm.
Given:
$s - a = 8 cm, s - b = 7 cm and s - c = 5 cm$
By adding all these, we get:
$s - a + s - b + s - c = 8 + 7 + 5 \Rightarrow 3 s - (a + b + c) = 20 [2 s = (a + b + c)] \Rightarrow 3 s - 2 s = 20 \Rightarrow s = 20 cm$

$By applying Heron' s formula, we get : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{20 \times 8 \times 7 \times 5} = \sqrt{5 \times 4 \times 7 \times 5 \times 4 \times 2} = 5 \times 4 \times \sqrt{14} = 20 \sqrt{14} {cm}^{2}$

Page No 242:

Question 18:

For an isosceles right-angled triangle having each of equal sides 'a', we have
I. $Area = \frac{1}{2} a^{2}$
II. $Perimeter = (2 + \sqrt{2}) a$
III. Hypotenuse = 2a
Which of the following is true?
(a) I only
(b) II only
(c) I and II
(d) I and III

Answer:

(c) I and II
Area of triangle = $\frac{1}{2} \times Base \times Height = \frac{1}{2} \times a \times a = \frac{1}{2} a^{2}$
Hypotenuse = $\sqrt{a^{2} + a^{2}} = \sqrt{2 a^{2}} = \sqrt{2} a$
Perimeter = $a + a + \sqrt{2} a = 2 a + \sqrt{2} a = a (2 + \sqrt{2})$

Page No 242:

Question 19:

For an isosceles triangle having base b and each of the equal sides as a, we have:
I. $Area = \frac{b \cdot \sqrt{4 a^{2} - b^{2}}}{4}$
II. Perimeter = (2a + b)
III. $Height = \frac{1}{2} \sqrt{4 a^{2} - b^{2}}$
Which of the following is true?
(a) I only
(b) I and II only
(c) II and III only
(d) I, II and III

Answer:

(d) I, II and III
For an isosceles triangle having base as b and each of the equal sides as a, we have:
I. $Area = \frac{b \sqrt{4 a^{2} - b^{2}}}{4}$
II. Perimeter = (2a + b)
III. $Height = \frac{1}{2} \sqrt{4 a^{2} - b^{2}}$

Page No 242:

Question 20:

Assertion: Area of an equilateral triangle having each side equal to 4 cm is $4 \sqrt{3} sq$ cm.
Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4} a^{2}$ sq units.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (4)^{2} = \frac{\sqrt{3}}{4} \times 16 = 4 \sqrt{3} {cm}^{2}$
Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4} a^{2}$ sq. units.
Hence, it is true.

Page No 243:

Question 21:

Assertion: The area of an isosceles triangle having base = 8 cm and each of the equal sides = 5 cm is 12 cm².
Reason: The area of an isosceles triangle having each of the equal sides as a and base = b is $\frac{1}{4} b \sqrt{4 a^{2} - b^{2}}$ .
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 5 cm and b = 8 cm Thus, we have : \frac{8}{4} \times \sqrt{4 (5)^{2} - 8^{2}} = 2 \times \sqrt{100 - 64} = 2 \times \sqrt{36} = 2 \times 6 = 12 {cm}^{2}$

Hence, Assertion is true.

Reason:
Area of an isosceles triangle having each of the equal sides as a and base as b is $\frac{1}{4} b \sqrt{4 a^{2} - b^{2}}$ .
Hence, it is true.

Page No 243:

Question 22:

Assertion: The area of an equilateral triangle having side 4 cm is 3 cm².
Reason: The area of an equilateral triangle having each side a is $(\frac{\sqrt{3}}{4} a^{2})$ sq units.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) Assertion is false and Reason is true.

Assertion:
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (4)^{2} = \frac{\sqrt{3}}{4} \times 16 = 4 \sqrt{3} {cm}^{2}$
Hence, Assertion is false.
Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4} a^{2}$ square units. Hence, it is true.

Page No 243:

Question 23:

Assertion: The sides of a ∆ABC are in the ratio 2 : 3 : 4 and its perimeter is 36 cm. Then, $ar (∆ A B C) = 12 \sqrt{15} {cm}^{2}$ .
Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then its area $= \sqrt{(s - a) (s - b) (s - c)}$ .
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(c) Assertion is true and Reason is false.
Assertion:
Let the sides of the triangle be 2x cm, 3x cm and 4x cm.
Perimeter = Sum of all sides
or, 36 = 2x + 3x + 4x
or, 9x = 36
or, x = 4
Thus, the sides of the triangle are 2 $\times$ 4 cm, 3 $\times$ 4 cm and 4 $\times$ 4 cm, i.e., 8 cm, 12 cm and 16 cm.

Now,
$Let : a = 8 cm, b = 12 cm and c = 16 cm s = \frac{36}{2} = 18 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{18 (18 - 8) (18 - 12) (18 - 16)} = \sqrt{18 \times 10 \times 6 \times 2} = \sqrt{6 \times 3 \times 5 \times 2 \times 6 \times 2} = 6 \times 2 \sqrt{15} = 12 \sqrt{15} {cm}^{2}$
Hence, Assertion is true.
Reason: If 2s = (a + b + c), where a, b and c are the sides of a triangle, then its area $= \sqrt{(s - a) (s - b) (s - c)}$ .
Hence, it is false.

Area should be $\sqrt{s (s - a) (s - b) (s - c)}$ .

Page No 243:

Question 24:

Assertion: The area of an isosceles triangle having base = 24 cm and each of the equal sides equal to 13 cm is 60 cm².
Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then area $= \sqrt{s (s - a) (s - b) (s - c)}$ .
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Assertion:
$Let : a = 24 cm, b = 13 cm and c = 13 cm s = \frac{a + b + c}{2} = \frac{24 + 13 + 13}{2} = 25 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{25 (25 - 24) (25 - 13) (25 - 13)} = \sqrt{25 \times 1 \times 12 \times 12} = \sqrt{5 \times 5 \times 12 \times 12} = 5 \times 12 = 60 {cm}^{2}$

Reason:
If 2s = (a + b + c), where a, b and c are the sides of a triangle, then area $= \sqrt{s (s - a) (s - b) (s - c)}$ .

Page No 243:

Question 25:

If the base of an isosceles triangle is 6 cm and its perimeter is 16 cm, then its area is 12 cm².

Answer:

True
Let the equal sides of the isosceles triangle be a cm each.
We have:
Base of the triangle, b = 6 cm
Perimeter = 16 cm
or, a + a + 6 = 16
or, 2a = 10
or, a = 5 cm

Now,
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 5 cm and b = 6 cm Thus, we have : \frac{6}{4} \times \sqrt{4 (5)^{2} - 6^{2}} = 1.5 \times \sqrt{100 - 36} = 1.5 \times \sqrt{64} = 1.5 \times 8 = 12 {cm}^{2}$

Page No 243:

Question 26:

If each side of an equilateral triangle is 8 cm long, then its area is $20 \sqrt{3}$ cm².

Answer:

False
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (8)^{2} = \frac{\sqrt{3}}{4} \times 64 = 16 \sqrt{3} {cm}^{2}$

Page No 243:

Question 27:

If the sides of a triangular field measure 51 m, 37 m and 20 m, then the cost of levelling it at Rs 5 per m² is Rs 1530.

Answer:

True

$Let : a = 51 m, b = 37 m and c = 20 m s = \frac{a + b + c}{2} = \frac{51 + 37 + 20}{2} = 54 m By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{54 (54 - 51) (54 - 37) (54 - 20)} = \sqrt{54 \times 3 \times 17 \times 34} = \sqrt{3 \times 3 \times 3 \times 2 \times 3 \times 17 \times 17 \times 2} = 17 \times 2 \times 3 \times 3 = 306 m^{2}$
Cost of levelling 1 m² of area = Rs 5
Cost of levelling 306 m² of area = $5 \times 306 = Rs 1530$

Page No 244:

Question 28:

Match the following columns.

Column I	Column II
(a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm.	(p) 6
(b) The area of an equilateral triangle is $4 \sqrt{3}$ cm². The perimeter of the triangle is ...... cm.	(q) 4
(c) If the height of an equilateral triangle is $3 \sqrt{3}$ cm, then each side of the triangle measures ...... cm.	(r) 24
(d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm.	(s) 12

(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer:

(a)
$Let : a = 26 cm, b = 28 cm and c = 30 cm s = \frac{a + b + c}{2} = \frac{26 + 28 + 30}{2} = 42 cm By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{42 (42 - 26) (42 - 28) (42 - 30)} = \sqrt{42 \times 12 \times 16 \times 14} = \sqrt{14 \times 3 \times 2 \times 3 \times 2 \times 4 \times 4 \times 14} = 14 \times 4 \times 2 \times 3 = 336 {cm}^{2}$

Hence, the height of the triangle corresponding to 28 cm is given by:
$Area of triangle = 336 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 336 \Rightarrow Height = \frac{336 \times 2}{28} = 24 cm$

(b)
$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} \Rightarrow \frac{\sqrt{3}}{4} \times (Side)^{2} = 4 \sqrt{3}$
$\Rightarrow {(Side)}^{2} = 16 \Rightarrow Side = 4 cm$

Perimeter = 3 × Side = 3 × 4 = 12 cm

(c)
$Height = \frac{\sqrt{3}}{2} \times Side \Rightarrow Side = \frac{2 \times Height}{\sqrt{3}} = \frac{2 \times 3 \sqrt{3}}{\sqrt{3}} = 6 cm$

(d)
$Height of isosceles triangle = \frac{1}{2} \sqrt{4 a^{2} - b^{2}} = \frac{1}{2} \sqrt{4 {(5)}^{2} - 6^{2}} (a = 5 cm and b = 6 cm) = \frac{1}{2} \times \sqrt{100 - 36} = \frac{1}{2} \times \sqrt{64} = \frac{1}{2} \times 8 = 4 cm$

Column I	Column II
(a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm.	(r) 24
(b) The area of an equilateral triangle is $4 \sqrt{3}$ cm². The perimeter of the triangle is ...... cm.	(s) 12
(c) If the height of an equilateral triangle is $3 \sqrt{3}$ cm, then each side of the triangle measures ...... cm.	(p) 6
(d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm.	(q) 4

Page No 244:

Question 29:

A park in the shape of a quadrilateral ABCD has AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and ∠C = 90°. Find the area of the park.

Answer:

We know that $△$ BCD is a right triangle.
∴ $B D = \sqrt{B C^{2} + C D^{2}} = \sqrt{12^{2} + 5^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13 m$

$Now, Area of triangle B C D = \frac{1}{2} \times base \times height = \frac{1}{2} \times B C \times C D = \frac{1}{2} \times 12 \times 5 = 30 m^{2}$

$Let : a = 9 m, b = 8 m and c = 13 m s = \frac{a + b + c}{2} = \frac{9 + 8 + 13}{2} = 15 m By Heron' s formula, we have : Area of triangle A B D = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{15 (15 - 9) (15 - 8) (15 - 13)} = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{5 \times 3 \times 3 \times 2 \times 7 \times 2} = 3 \times 2 \sqrt{35} = 6 \sqrt{35} = 6 \times 5.9 = 35.4 m^{2}$
Now,
Area of quadrilateral ABCD = Area of $△$ BCD + Area of $△$ ABD
= (30 + 35.4) m² = 65.4 m²

Page No 244:

Question 30:

Find the area of a parallelogram ABCD in which AB = 60 cm, BC = 40 cm and AC = 80 cm.

Answer:

$Let : a = 40 cm, b = 80 cm and c = 60 cm s = \frac{a + b + c}{2} = \frac{40 + 80 + 60}{2} = 90 cm By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{90 (90 - 40) (90 - 80) (90 - 60)} = \sqrt{90 \times 50 \times 10 \times 30} = \sqrt{30 \times 3 \times 10 \times 5 \times 10 \times 30} = 30 \times 10 \sqrt{15} = 300 \times 3.87 = 1161 {cm}^{2}$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2 \times 1161 {cm}^{2} = 2322 {cm}^{2}$

Page No 245:

Question 31:

A piece of land is in the shape of a rhombus ABCD in which each side measures 100 m and diagonal AC is 160 m long. Find the area of the rhombus.

Answer:

$Let : a = 100 m, b = 160 m and c = 100 m s = \frac{a + b + c}{2} = \frac{100 + 160 + 100}{2} = 180 m By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{180 (180 - 100) (180 - 160) (180 - 100)} = \sqrt{180 \times 80 \times 20 \times 80} = \sqrt{20 \times 3 \times 3 \times 80 \times 20 \times 80} = 20 \times 3 \times 80 = 4800 m^{2}$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2 \times 4800 = 9600 m^{2}$

Page No 245:

Question 32:

A floral design on a floor is made up of 16 triangular tiles, each having sides 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of Rs 2.50 per cm².

Answer:

Area of one triangular tile can be found in the following manner:
$Let : a = 9 cm, b = 28 cm and c = 35 cm s = \frac{a + b + c}{2} = \frac{9 + 28 + 35}{2} = 36 cm By Heron' s formula, we have : Area of triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{36 (36 - 9) (36 - 28) (36 - 35)} = \sqrt{36 \times 27 \times 8 \times 1} = \sqrt{9 \times 4 \times 3 \times 9 \times 4 \times 2} = 9 \times 4 \sqrt{6} = 88.2 {cm}^{2}$

∴ Area of 16 triangular tiles = $16 \times 88.2 = 1411.2 {cm}^{2}$
Cost of polishing 1 cm² of area = Rs 2.5
Cost of polishing 1411.2 cm² of area = $2.5 \times 1411.2 = Rs 3528$

Page No 245:

Question 33:

A like in the shape of a square with each diagonal 32 cm and having a tail in the shape on an isosceles triangle of base 8 cm and each side 6 cm, is made of three different shades as shown in the figure. How much paper of each shade has been used in it?

Answer:

$Every square is a rhombus . Area of rhombus = \frac{1}{2} (I^{st} diagonal \times {II}^{nd} diagonal) Here, each diagonal measures 32 cm . Therefore, we can say : Area of square A B C D = \frac{1}{2} \times (Diagonal)^{2} = \frac{1}{2} \times 32 \times 32 = 512 {cm}^{2}$

We know that a diagonal of a parallelogram divides into two triangles of equal areas. Also, a square is a parallelogram. Therefore, we can say that a diagonal of a square divides into two triangles of equal areas.
∴ $Area of △ A B D = Area of △ B D C = \frac{1}{2} \times 512 = 256 {cm}^{2}$
$Area of isosceles △ C E F = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 6 cm and b = 8 cm Thus, we have : \frac{8}{4} \times \sqrt{4 (6)^{2} - 8^{2}} = \frac{8}{4} \times \sqrt{144 - 64} = \frac{8}{4} \times \sqrt{80} = \frac{8}{4} \times 4 \sqrt{5} = 8 \sqrt{5} = 8 \times 2.24 = 17.92 {cm}^{2}$

Page No 250:

Question 1:

Each side of an equilateral triangle is 8 cm. Its altitude is
(a) $2 \sqrt{2} cm$
(b) $2 \sqrt{3} cm$
(c) $4 \sqrt{3} cm$
(d) $2 \sqrt{6} cm$

Answer:

Page No 250:

Question 2:

The perimeter of an isosceles right-angled triangle having a as each of the equal sides is
(a) $(1 + \sqrt{2}) a$
(b) $(2 + \sqrt{2}) a$
(c) $3 a$
(d) $(3 + \sqrt{2}) a$

Answer:

(b) $(2 + \sqrt{2}) a$
For an isosceles right-angled triangle having each equal side as a, we have:
Hypotenuse = $\sqrt{a^{2} + a^{2}} = \sqrt{2 a^{2}} = \sqrt{2} a$
Perimeter = $a + a + \sqrt{2} a = 2 a + \sqrt{2} a = (2 + \sqrt{2}) a$

Page No 250:

Question 3:

For an isosceles triangle having base = 12 cm and each of the equal sides equal to 10 cm, the height is
(a) 12 cm
(b) 16 cm
(c) 6 cm
(d) 8 cm

Answer:

(d) 8 cm
$Height of isosceles triangle = \frac{1}{2} \sqrt{4 a^{2} - b^{2}} = \frac{1}{2} \sqrt{4 {(10)}^{2} - 12^{2}} (a = 10 cm and b = 12 cm) = \frac{1}{2} \times \sqrt{400 - 144} = \frac{1}{2} \times \sqrt{256} = \frac{1}{2} \times 16 = 8 cm$

Page No 250:

Question 4:

Find the area of an equilateral triangle having each side 6 cm.

Answer:

$Area of equilateral triangle = \frac{\sqrt{3}}{4} \times (Side)^{2} = \frac{\sqrt{3}}{4} \times (6)^{2} = \frac{\sqrt{3}}{4} \times 36 = 9 \sqrt{3} {cm}^{2}$

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Question 5:

Using Heron's formula find the area of ∆ABC in which BC = 13 cm, AC = 14 cm and AB = 15 cm.

Answer:

$Let : a = 13 cm, b = 14 cm and c = 15 cm s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21 cm By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{21 (21 - 13) (21 - 14) (21 - 15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 3 \times 2} = 7 \times 3 \times 2 \times 2 = 84 {cm}^{2}$

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Question 6:

The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.

Answer:

Let the sides of the triangle be 13x cm, 14x cm and 15x cm.
Perimeter = Sum of all sides
or, 84 = 13x + 14x + 15x
or, 42x = 84
or, x = 2
So, sides of the triangle are 13 $\times$ 2, 14 $\times$ 2 and 15 $\times$ 2, i.e., 26 cm, 28 cm and 30 cm.
$Let : a = 26 cm, b = 28 cm and c = 30 cm s = \frac{84}{2} = 42 c m By Heron' s formula, we have : Area of the triangle = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{42 (42 - 26) (42 - 28) (42 - 30)} = \sqrt{42 \times 16 \times 14 \times 12} = \sqrt{14 \times 3 \times 4 \times 4 \times 14 \times 3 \times 2 \times 2} = 14 \times 3 \times 4 \times 2 = 336 {cm}^{2}$

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Question 7:

Find the area of ∆ABC in which BC = 8 cm, AC = 15 cm and AB = 17 cm. Find the length of altitude drawn on AB.

Answer:

$Let : a = 8 cm, b = 15 cm and c = 17 cm s = \frac{a + b + c}{2} = \frac{8 + 15 + 17}{2} = 20 cm By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{20 (20 - 8) (20 - 15) (20 - 17)} = \sqrt{20 \times 12 \times 5 \times 3} = \sqrt{4 \times 5 \times 4 \times 3 \times 5 \times 3} = 4 \times 5 \times 3 = 60 {cm}^{2}$

Altitude of the triangle corresponding to AB:
$Area of triangle = 60 {cm}^{2} \Rightarrow \frac{1}{2} \times Base \times Height = 60 \Rightarrow Height = \frac{60 \times 2}{17} = 7.06 cm$

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Question 8:

An isosceles triangle has perimeter 30 cm and each of its equal sides is 12 cm. Find the area of the triangle.

Answer:

Let the base of the isosceles triangle be b cm.
Now,
Equal side of the triangle = 12 cm

Perimeter = 30 cm
or, 12 + 12 + b = 30
or, 24 + b = 30
or, b = 6 cm
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 12 cm and b = 6 cm Thus, we have : = \frac{6}{4} \times \sqrt{4 (12)^{2} - 6^{2}} = 1.5 \times \sqrt{576 - 36} = 1.5 \times \sqrt{540} = 1.5 \times \sqrt{6 \times 6 \times 15} = 1.5 \times 6 \sqrt{15} = 9 \sqrt{15} {cm}^{2}$

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Question 9:

The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3 : 2. Find the area of the triangle.

Answer:

Let the equal sides of the isosceles triangle (a) be 3x cm and base of the triangle (b) be 2x cm.

(i) Perimeter = 32 cm
or, 3x + 3x + 2x = 32
or, 8x = 32
or, x = 4

∴ Equal sides of the triangle = 3x = 3(4) = 12 cm
Base of the triangle = 2x = 2(4) = 8 cm
Now,
$Area of isosceles triangle = \frac{b}{4} \sqrt{4 a^{2} - b^{2}} Here, a = 12 cm and b = 8 cm Thus, we have : = \frac{8}{4} \times \sqrt{4 (12)^{2} - 8^{2}} = 2 \times \sqrt{576 - 64} = 2 \times \sqrt{512} = 2 \times \sqrt{256 \times 2} = 2 \times 16 \sqrt{2} = 32 \sqrt{2} {cm}^{2}$

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Question 10:

Given a ∆ABC in which
I. ∠A, ∠B and ∠C are in the ratio 3 : 2 : 1.
II. AB, AC and BC are in the ratio $3 : \sqrt{3} : 2 \sqrt{3}$ and $A B = 3 \sqrt{3} cm$ .
Is ∆ABC a right triangle?
The question give above has two Statements I and II. Answer the question by using instructions given below:
(a) If the question can be answered by one of the given statements only and not by the other.
(b) If the question can be answered by using either statement along.
(c) If the question can be answered by using both the statements but cannot be answered by using either statement.
(d) If the question cannot be answered even by using both the statements together.

Answer:

(b)
Let $∠$ A, $∠$ B and $∠$ C be 3x, 2x and x, respectively.
$∴ 3 x + 2 x + x = 180 ° (Angle sum property) \Rightarrow 6 x = 180 ° \Rightarrow x = 30 °$
$Now, we have : ∠ A = 3 x = 3 \times 30 ° = 90 ° ∠ B = 2 x = 2 \times 30 ° = 60 ° ∠ C = x = 30 °$

Hence, it is a right triangle at A.
Let the sides of the triangle be $3 x cm, \sqrt{3} x cm and 2 \sqrt{3} x cm .$
Also,
$A B = 3 \sqrt{3} cm or, 3 x = 3 \sqrt{3} or, x = 3$
Now,
$A C = \sqrt{3} x = \sqrt{3} \times \sqrt{3} = 3 cm B C = 2 \sqrt{3} x = 2 \sqrt{3} \times \sqrt{3} = 6 cm$

$B C^{2} = 6^{2} = 36 cm A B^{2} + A C^{2} = {(3 \sqrt{3})}^{2} + {(3)}^{2} = 27 + 9 = 36 cm Here, B C^{2} = A B^{2} + A C^{2} Thus, it is a right - angled triangle .$

So, it can be answered using either statement.

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Question 11:

In the given figure ∆ABC and ∆DBC have the same base BC such that AB = 120 m, AC = 122 m, BC = 22 m, BD = 24 m and CD = 26 m. Find the area of the shaded region.

Answer:

$Let : a = 22 m, b = 122 m and c = 120 m s = \frac{a + b + c}{2} = \frac{22 + 122 + 120}{2} = 132 m By Heron' s formula, we have : Area of triangle A B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{132 (132 - 22) (132 - 122) (132 - 120)} = \sqrt{132 \times 110 \times 10 \times 12} = \sqrt{11 \times 12 \times 11 \times 10 \times 10 \times 12} = 11 \times 12 \times 10 = 1320 m^{2}$

Now,
$Let : a = 22 m, b = 24 m and c = 26 m s = \frac{a + b + c}{2} = \frac{22 + 24 + 26}{2} = 36 m By Heron' s formula, we have : Area of triangle O B C = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{36 (36 - 22) (36 - 24) (36 - 26)} = \sqrt{36 \times 14 \times 12 \times 10} = \sqrt{12 \times 3 \times 7 \times 2 \times 12 \times 5 \times 2} = 12 \times 2 \sqrt{105} = 24 \sqrt{105} = 246 m^{2}$

$∴ Area of the shaded portion = Area of ∆ A B C - Area of ∆ O B C = (1320 - 246) m^{2} = 1074 m^{2}$

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Question 12:

A point O is taken inside an equilateral ∆ABC. If OL ⊥ BC, OM ⊥ AC and ON ⊥ AB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of ∆ABC.

Answer:

$Let each side of △ A B C b e x cm . ∴ a r (∆ A B C) = a r (∆ B O C) + ar (∆ A O C) + ar (∆ A O B) = \frac{1}{2} \times B C \times O L + \frac{1}{2} \times A C \times O M + \frac{1}{2} \times A B \times O N = \frac{1}{2} \times x \times 14 + \frac{1}{2} \times x \times 10 + \frac{1}{2} \times x \times 6 = \frac{1}{2} x (14 + 10 + 6) = \frac{1}{2} x \times 30 = 15 x {cm}^{2} a r (∆ A B C) = 15 x {cm}^{2} or, \frac{\sqrt{3}}{4} x^{2} = 15 x or, \sqrt{3} x^{2} = 60 x or, \sqrt{3} x^{2} - 60 x = 0 or, \sqrt{3} x (x - 20 \sqrt{3}) = 0 or, \sqrt{3} x = 0 a n d (x - 20 \sqrt{3}) = 0 or, x = 0 (Side cannnot be 0 .) and, x = 20 \sqrt{3} ∴ ar (∆ A B C) = 15 x = 15 \times 20 \sqrt{3} = 300 \sqrt{3} {cm}^{2}$

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