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#### Question 1:

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

We have:
Base = 24 cm
Height = 14.5 cm

Now,

#### 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.

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle =
We have:

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

#### 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.

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:

#### 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.

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:

#### 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.

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.

#### 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.

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$×$5 = 25 m
12$×$5 = 60 m
13$×$5 = 65 m

Now,

#### 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.

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$×$10 = 250 m
17$×$10 = 170 m
12$×$10 = 120 m

Now,

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 =

#### 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.

(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:

#### Question 9:

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

#### Question 10:

The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle. Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,

Also,

∴ Perimeter = 80 + 41 + 41  = 162 cm

#### 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.

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

So, equal sides of the triangle are 12 cm each.
Also,
Base = $\frac{3}{2}$a =
(ii)

(iii)

#### Question 12:

If the area of an equilateral triangle is , find its perimeter.

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

#### Question 13:

If the area of an equilateral triangle is , find its height.

Now, we have:

#### Question 14:

The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle. Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,

#### 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 $\sqrt{3}=1.732$.

Side of the equilateral triangle = 8 cm

(i)

(ii)

#### Question 16:

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

Height of the equilateral triangle = 9 cm
Thus, we have:

Also,

#### 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. We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.

Now,
Area of 1 triangular piece of cloth = 490 cm2
Area of 12 triangular pieces of cloth =

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

Now,
Area of 16 triangular-shaped tiles =
Cost of polishing tiles of area 1 cm2 = Rs 1
Cost of polishing tiles of area 1536 cm2 =

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

∴

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

#### 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°. We know that $△$DBC is a right-angled triangle.

Now,

∴ 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

#### 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: $\sqrt{3}=1.73\right]$] We know that $△$BAD is a right-angled triangle.

∴

Also, we know that $△$BDC is an equilateral triangle.

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

#### Question 22:

Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm. We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### Question 23:

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: $\sqrt{3}=1.73$] We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### Question 24:

In the given figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL BD and CMBD such that AL = 16.8 cm and CM = 13.2 cm. Calculate the area of quadrilateral ABCD. #### 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)
(d) 45 cm2

(b) 30 cm2

#### 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

(a) 96 cm2

#### Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the triangle is
(a)
(b)
(c)
(d) 48 cm2

(b)

#### 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)
(b)
(c)
(d)

(b)

#### 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)
(c) 4 cm
(d)

(c) 4 cm

#### Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a)
(b) 50 cm2
(c)
(d)
75 cm2

(b) 50 cm2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:

#### Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the triangle is
(a)
(b)
(c)
(d)
5 cm

(b)

#### Question 8:

The height of an equilateral triangle is 6 cm. Its area is
(a)
(b)
(c)
(d)
18 cm2

(a)

#### 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

(c) 384 m2

#### 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

(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$×$5 cm, 12$×$5 cm and 13$×$5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,

#### 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

(a) 24 cm

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:

#### 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

(b) 36 cm Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,

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

#### Question 13:

The area of an equilateral triangle is . Its perimeter is
(a) 36 cm
(b)
(c) 24 cm
(d) 30
cm

(a) 36 cm

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

#### 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

(c) 60 cm2

#### 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

(c) 336 cm2 Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,

#### Question 16:

The area of an equilateral triangle is . Its height is
(a)
(b)
(c)
(d) 9
cm

(a)

#### 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)
(b)
(c)
(d) 140 cm2

(c)
Let the sides of the triangle be a cm, b cm and c cm and the semi-perimeter be s cm.
Given:

By adding all these, we get:

#### Question 18:

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

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

#### Question 19:

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

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

#### Question 20:

Assertion: Area of an equilateral triangle having each side equal to 4 cm is 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.

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

Assertion:

Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4}{a}^{2}$ sq. units.
Hence, it is true.

#### 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.

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

Assertion:

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.

#### 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 $\left(\frac{\sqrt{3}}{4}{a}^{2}\right)$ 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.

(d) Assertion is false and Reason is true.

Assertion:

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.

#### Question 23:

Assertion: The sides of a ABC are in the ratio 2 : 3 : 4 and its perimeter is 36 cm. Then, .
Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then its area $=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.
(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.

(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$×$4 cm, 3$×$4 cm and 4$×$4 cm, i.e., 8 cm, 12 cm and 16 cm.

Now,

Hence, Assertion is true.
Reason: If 2s = (a + b + c), where a, b and c are the sides of a triangle, then its area $=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.
Hence, it is false.

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

#### 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 $=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.
(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.

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

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

#### Question 25:

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

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,

#### Question 26:

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

False

#### 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.

True

Cost of levelling 1 m2 of area = Rs 5
Cost of levelling 306 m2 of area =

#### 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) ......,

(a)

Hence, the height of the triangle corresponding to 28 cm is given by:

(b)

Perimeter = 3 × Side = 3 × 4 = 12 cm

(c)

(d)

 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

#### 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. We know that $△$BCD is a right triangle.
∴

Now,
Area of quadrilateral ABCD  = Area of $△$BCD + Area of $△$ABD
= (30 + 35.4) m2 = 65.4 m2

#### Question 30:

Find the area of a parallelogram ABCD in which AB = 60 cm, BC = 40 cm and AC = 80 cm. We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### 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. We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### 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. Area of one triangular tile can be found in the following manner:

∴ Area of 16 triangular tiles =
Cost of polishing 1 cm2 of area = Rs 2.5
Cost of polishing 1411.2 cm2 of area =

#### 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? 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.
∴

#### Question 1:

Each side of an equilateral triangle is 8 cm. Its altitude is
(a)
(b)
(c)
(d)

(c)

#### Question 2:

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

(b) $\left(2+\sqrt{2}\right)a$
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 = $a+a+\sqrt{2}a=2a+\sqrt{2}a=\left(2+\sqrt{2}\right)a$

#### 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

(d) 8 cm

#### Question 4:

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

#### Question 5:

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

#### 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.

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$×$2, 14$×$2 and 15$×$2, i.e., 26 cm, 28 cm and 30 cm.

#### 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.

Altitude of the triangle corresponding to AB:

#### Question 8:

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

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

#### 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.

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,

#### 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 .
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. (b)
Let $\angle$A, $\angle$ B and $\angle$C be 3x, 2x and x, respectively.

Hence, it is a right triangle at A.
Let the sides of the triangle be
Also,

Now,

So, it can be answered using either statement.

#### 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. Now,

#### Question 12:

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