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.

#### Question 1:

Draw a line segment AB = 5 cm and draw its perpendicular bisector. Steps of construction:
1. Draw a line segment AB = 5 cm.
2. With A as the centre and a radius equal to more than half of AB, draw two arcs: one above AB and the other below AB.
3. With B as the centre and the same radius same as above, draw two arcs cutting the previously drawn arcs at M and N, respectively.
4. Join MN intersecting AB at P.
Thus, MN perpendicularly bisects AB at P.

#### Question 2:

Draw an angle of 45° using scale and compasses only. Draw the bisector of this angle. Steps of construction:
1. Draw a line segment AB.
2. Draw an angle $\angle$BAC = 90$°$.
3. With A as the centre and a small radius, draw an arc cutting AB at P and AC at T.
4. With P as the centre and a radius more than half of PT, draw an arc.
5. With T as the centre and the same radius as above, draw another arc cutting the previously drawn arc at N .
6. Join AN cutting the arc at Z and produced it to X to make the ray AX.
7. With P as the centre and a radius more than half of PZ, draw an arc.
8. With Z as the centre and the same radius as above, draw another arc cutting the previously drawn arc at M .
9. Join AM.
Thus, AM bisects $\angle$BAX.

#### Question 3:

Construct an angle of 90° and draw its bisector. Steps of construction:
1. Draw a line segment AB.
2. With A as the centre and and a small radius, draw an  arc cutting AB at M.
3. With M as the centre and the same radius as above, draw an arc cutting the previously drawn arc at N.
4. With N as the centre and the same radius as above, draw an arc cutting the previously drawn arc at P.
5. Again, with N as the centre and a radius more than half of PN, draw an arc.
6. With P as the centre and the same radius as above, draw an arc cutting the previously drawm arc at Q.
7. Join AQ cutting the arc at O and produced it to C.
8. With O as the centre and a radius more than half of OM, draw an arc.
9. With M as the centre and the same radius as above, draw another arc cutting the previously drawn arc at point R.
10. Join AR.
Thus, AR bisects $\angle$BAC.

#### Question 4:

Construct an equilateral triangle each of  whose sides measures 5 cm. Steps of construction:
1. Draw a line segment AB = 5 cm.
2. With A as the centre and a radius equal to AB, draw an arc.
3. With B as the centre and the same radius as above, draw another arc cutting the previously drawn arc at C.
4. Join AC and BC.
Thus, $△$ABC is the required triangle.

#### Question 5:

Construct an equilateral triangle each of whose altitudes measures 5.4 cm. Steps of construction:
1. Draw a line XY.
2. Mark any point P.
3. From P draw PQ $\perp$ XY.
4. From P, set off PA = 5.4 cm, cutting PQ at A.
5. Construct $\angle$PAB = 30$°$ and $\angle$PAC = 30$°$, meeting XY at B and C, respectively.

Thus, $△$ABC is the required triangle.

#### Question 6:

Construct a ABC in which BC = 5 cm, AB = 3.8 cm and AC = 2.6 cm. Steps of construction:
1. Draw a line segment BC = 5 cm.
2. With B as the centre and a radius equal to 3.8 cm, draw an arc.
3. With C as the centre and a radius equal to 2.6 cm, draw an arc cutting the previously drawn arc at A.
4. Join AB and AC.
Thus, ABC is the required triangle.

#### Question 7:

Construct a ABC in which BC = 4.7 cm, ∠B = 60° and ∠C = 30°. Measure ∠A. Steps of construction:
1. Draw a line segment BC = 4.7 cm.
2. Construct $\angle$CBX = 60$°$.
3. Construct $\angle$BCY = 30$°$.
4. The ray BX and CY intersect at A.
Thus, $△$ABC is the required triangle.
When we measure $\angle$A, we get $\angle$A = 90$°$.

#### Question 8:

Construct an isosceles PQR whose base measures 5 cm and each of equal sides measures 4.5 cm. Steps of construction:
1. Draw a line segment PQ = 5 cm.
2. With P as the centre and radius equal to 4.5 cm, draw an arc.
3. With Q as the centre and the same radius as above, draw another arc cutting the previously drawn arc at R.
4. Join PR and QR.
Thus, PQR is the required triangle.

#### Question 9:

Construct an isosceles triangle whose base is 4.8 cm and whose vertical angle is 80°. Steps of construction:
1. Draw a line segment BC = 4.8 cm.
2. Construct $\angle$CBX = 80$°$.
4. Make $\angle$XBY = 90$°$.
5. Draw the right bisector PQ of BC intersecting BY at O.
6. With O as the centre and radius OB, draw a arc intersecting PQ at A.
7. Join AC and AB.
Thus, ABC is the required triangle.

#### Question 10:

Construct a right-angled triangle whose hypotenuse measures 5.3 cm and the length of one of whose sides containing the right angle measures 4.5 cm. Steps of construction:
1. Draw a line segment AB = 5.3 cm.
2. Find the midpoint O of BC.
3. With O as the centre and radius OB, draw a semicircle on BC.
4. With B as the centre and radius equal to 4.5 cm, draw an arc cutting the semicircle at A.
4. Join AB and AC.
Thus, ABC is the required triangle.

#### Question 11:

Construct a ABC in which ∠B = 30°, ∠C = 60° and the length of the perpendicular from the vertex A to the base BC is 4.8 cm. Steps of construction:
1. Draw a line segment XY.
2. Take a point D on XY and draw PD $\perp$ XY.
3. Along PD, set off DA = 4.8 cm.
4. Draw a line LM $\parallel$ XY.
5. Draw $\angle$LAB = 30$°$ and $\angle$MAC = 60$°$, meeting XY at B and C, respectively.
Thus, ABC is the required triangle.

#### Question 12:

Construct a PQR whose perimeter is 12 cm and the lengths of whose sides are in the ratio 3 : 2 : 4. 1. Draw a line segment XY = 12 cm.
2. In the downward direction, construct an acute angle with XY at X.
3. From X, set off (3 + 2 + 4) = 9 arcs of equal distances along XZ.
4. Mark points L, M and N such that XL = 3 units, LM = 2 units and MN = 4 units.
5. Join NY.
6. Through L and M, draw LQ $\parallel$ NY and MR $\parallel$ NY cutting XY at Q and R, respectively.
7. With Q as the centre and radius QX, draw an arc.
8. With R as the centre and radius RY, draw an arc, cutting the previously drawn arc at P.
9. Join PQ and PR.
Thus, $△$PQR is the required triangle.

#### Question 13:

Construct a ABC in which BC = 4.5 cm, ∠B = 60° and the sum of the other two sides is 8 cm. Steps of construction:
1. Draw a line segment BC = 4.5 cm.
2. Construct $\angle$CBX = 60$°$.
3. Set off BP = 8 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP at A.
6. Join AC.
Thus, $△$ABC is the required triangle.

#### Question 14:

Construct a ABC in which BC = 5.2 cm, ∠B = 30° and the difference of the other two sides is 3.5 cm. Steps of construction:
1. Draw a line segment BC = 5.2 cm.
2. Construct $\angle$CBX = 30$°$.
3. Set off BP = 3.5 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP produced at A.
6. Join AC.
Thus, $△$ABC is the required triangle.

#### Question 1:

Which of the following angles can be constructed using ruler and compass only?
(a) 25°
(b) 50°
(c) 22.5°
(d) 42.5°

(c) 22.5°

We can draw 45° with the help of a ruler and a compass and bisect it to get 22.5°.

#### Question 2:

Which of the following angles can be constructed using ruler and compass only?
(a) 65°
(b) 72°
(c) 80°
(d) 67.5°

(d) 67.5°
It is because we can draw (90°+ 45° = 135°) with the help of a ruler and a compass and bisect it to get 67.5°.

#### Question 3:

Which of the following angles cannot be constructed using ruler and compass only?
(a) 40°
(b) 12
(c) 135°
(d) 37.5°

(a) 40°
It is because we cannot draw an angle of 80° using a compass and a ruler (half of 80° is 40°).

#### Question 4:

Which of the following angles cannot be constructed using ruler and compass only?
(a) $22\frac{1}{2}°$
(b) 15°
(c) $52\frac{1}{2}°$
(d) $32\frac{1}{2}°$

(d) $32\frac{1}{2}°$
It is because we can not draw 65° using a compass and a ruler (half of 65° is $32\frac{1}{2}°$).

#### Question 5:

The construction of a ABC in which AB = 6 cm, ∠A = 45° is possible when (BC + AC) is
(a) 7 cm
(b) 5.8 cm
(c) 5 cm
(d) 4.9 cm

(a) 7 cm
We know that the sum of any two sides of a triangle is always greater than the third side.
i.e., BC + AC > AB
Here, BC + AC > 6 cm and 7 > 6

#### Question 6:

The construction of a PQR in which QR = 5.4 cm and ∠Q = 60° is not possible when (PQ + QR) is
(a) 6 cm
(b) 6.5 cm
(c) 5 cm
(d) 7 cm

(c) 5 cm
Given: QR = 5.4 cm
Hence, PQ + QR can not be less than 5.4.

#### Question 7:

The construction of a ABC in which AB = 7 cm, ∠A = 75° is possible when (BCAC) is equal to
(a) 7.5 cm
(b) 7 cm
(c) 8 cm
(d) 6.5 cm

(d) 6.5 cm
It is because the difference of any two sides of a triangle is always less than the third side.

i.e., BC $-$ AC < AB
Here, BC $-$ AC < 7 cm and 6.5 < 7.

#### Question 8:

The construction of a ABC in which BC = 6 cm and ∠B = 50° is not possible when (AB AC) is equal to
(a) 5.6 cm
(b) 5 cm
(c) 6 cm
(d) 4.8 cm

(c) 6 cm
It is because the difference of any two sides of a triangle is always less than the third side.
i.e., AB $-$ AC < BC
So, AB
$-$ AC < 6
Thus,
the construction of a ABC in which BC = 6 cm and ∠B = 50° is not possible when (AB AC) is equal to 6 cm.

#### Question 9:

Is it possible to construct a triangle whose sides measure 7 cm, 5 cm and 12 cm?
(a) Yes
(b) No

(b) No
It is because the sum of any two sides of a triangle is always greater than the third side.
Here, the sum of two sides (i.e., 7+5) is equal to the third side (12 cm).

#### Question 10:

Is it possible to construct a triangle whose sides measure 6 cm, 5 cm and 10 cm?
(a) Yes
(b) No

(a) Yes
It is because the sum of any two sides of a triangle is always greater than the third side.

#### Question 11:

Is it possible to construct a ABC in which BC = 5 cm, ∠B = 120° and ∠C = 60°?
(a) Yes
(b) No

(b) No
It is because the sum of two angles of a triangle is always less than 180°.
Here, the sum of is
180°.

#### Question 12:

Is it possible to construct a ABC in which ∠A = 60°, ∠B = 70° and ∠C = 60°?
(a) Yes
(b) No

(b) No
It is because, the sum of the angles of a triangle cannot be greater than 180°.
Here, the sum of is
190°, which is greater than 180°.

#### Question 13:

Is it possible to construct an angle of 35° using ruler and compass only?
(a) Yes
(b) No

(b) No
It is because we cannot draw an angle of 70° using a compass and a ruler (here, half of 70° is 35°).

#### Question 14:

Is it possible to construct an angle of 67.5° using ruler and compass only?
(a) Yes
(b) No

(a) Yes
We can draw an angle of 135° using a compass and a ruler and bisect it to get 67.5°.

#### Question 1:

Is it possible to construct a ABC in which BC = 5 cm, ∠B = 60°, ∠C = 60°?
(a) Yes
(b) No

(a) Yes
It is because the sum of two angles of a triangle is always less than 180°.
Here, the sum of is
120°, which is less than 180°.

#### Question 2:

Is it possible to construct a ABC in which AB = 5 cm, BC = 5 cm and AC = 10 cm?
(a) Yes
(b) No

(b) No
It is because the sum of any two sides of a triangle is always greater than the third side.
Here, the sum of two sides (i.e., 5+5 = 10 cm) is equal to the third side (10 cm).

#### Question 3:

Is it possible to construct an angle of 75° using ruler and compass only?
(a) Yes
(b) No

(a) Yes
We can draw 60° and 15° using a compass and a ruler.
Therefore, we can draw an angle of 75o.

#### Question 4:

Construct a ABC whose perimeter is 12 cm and whose base angle are 45° and 60°. Steps of construction:
1. Draw a line segment XY = 12 cm.
2. Construct an angle of 45$°$ and bisect it to get $\angle$YXP.
3. Construct an angle of 60$°$ and bisect it to get $\angle$XYQ.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AX and AY, cutting XY at B and C, respectively.
6. Join AB and AC.
Thus, $△$ABC is the required triangle.

#### Question 5:

Construct a ABC whose perimeter is 15 cm and sides are in the ratio 3 : 4 : 5. Steps of construction:
1. Draw a line segment XY = 15 cm.
2. Construct an acute angle with XY in the downward direction.
3. With X, set off (3 + 4 + 5) = 12 equal distances along XP.
4. Mark points L, M and N such that XL = 3 units, LM = 4 units and MN = 5 units.
5. Join YN.
6. Through L and M, draw LB $\parallel$ YN and MC $\parallel$ YN, cutting XY at B and C, respectively.
7. With B as the centre and radius XB, draw an arc.
8. With C as the centre and radius CY, draw an arc cutting the previously drawn arc at A.
9. Join AB and AC.
Thus, $△$ABC is the required triangle.

#### Question 6:

Construct an isosceles triangle whose base is 6 cm and whose vertical angle 18 75°. Steps of construction:
1. Draw a line segment BC = 6 cm.
2. Construct $\angle$CBX = 75$°$.
4. Make $\angle$XBY = 90$°$.
5. Draw the right bisector PQ of BC, intersecting BY at O.
6. With O as the centre and radius OB, draw a circle intersecting PQ at A.
7. Join AC and AB.
Thus, ABC is the required triangle.

#### Question 7:

Draw a right-angled triangle having hypotenuse = 6 cm and one of the sides containing the right angle having length 4.5 cm. Steps of construction:
1. Draw a line segment BC = 6 cm.
2. Find the mid point O of BC.
3. With O as the centre and a radius equal to OB, draw a semicircle on BC.
4. With B as the centre and a radius equal to 4.5 cm, draw an arc cutting the semicircle at A.
4. Join AB and AC.
Thus, $∆$ABC is the required triangle.

#### Question 8:

Construct a ABC in which ∠B = 60°, ∠C = 45° and the length of perpendicular from vertex A to base BC as 5 cm. Steps of construction:
1. Draw a line XY.
2. Take a point D on XY and draw PD $\perp$ XY.
3. Along PD, set off DA = 5 cm.
4. Draw a line LM $\parallel$ XY.
5. Draw $\angle$LAB = 60$°$ and $\angle$MAC = 45$°$ meeting XY at B and C, respectively.
Thus, $∆$ABC is the required triangle.

#### Question 9:

Construct an angle of $22\frac{1}{2}°$. Steps of construction:
1. Draw a ray AB.
2. Draw an angle $\angle$BAE = 45$°$.
3. With A as the centre and a small radius, draw an arc cutting AB at P and AE at Q.
4. With P as the centre and a radius more than half of PQ, draw an arc.
5. With Q as the centre and the same radius as above, draw another arc cutting the previously drawn arc at D.
Thus, $\angle$BAC is the required angle of measure 22.5o.

#### Question 10:

Construct an angle of 135°. Steps of construction:
1. Draw a line XY.
2. Take a point A on XY.
3. With A as centre, draw a semi circle, cutting XY at P and Q.
4. Construct $\angle$YAC = 90$°$.
5. Draw AB, bisector of $\angle$XAC.
Thus, $\angle$YAB = 135$°$

#### Question 11:

Construct an equilateral triangle of side 5 cm. Steps of construction:
1. Draw a line segment AB = 5 cm.
2. With A as the centre and a radius equal to AB, draw an arc.
3. With B as the centre and the same radius as above, draw another arc cutting the previously drawn arc at C.
4. Join AC and BC.
Thus, $△$ABC is the required triangle.

#### Question 12:

Construct a square of side 4 cm. Steps of construction:
1. Draw a line segment AB = 4 cm.
2. Construct $\angle$BAX = 90$°$ and $\angle$ABY = 90$°$.
3. Set off AD = 4 cm and BC = 4 cm.
4. Join DC.
Thus, $\square$ABCD is the required square.

#### Question 13:

Draw a line segment AB of length 5.2 cm and construct the perpendicularr bisector of AB. Steps of construction:
1. Draw a line segment AB = 5.2 cm.
2. With A as the centre and a radius more than half of AB, draw two arcs: one above AB and the other below AB.
3. With B as the centre and the same radius as above, draw two arcs, cutting the previously drawn arcs at points C and D.
4. Join CD, intersecting AB at P.
Thus, CD bisects AB perpendicularly at point P.

#### Question 14:

Construct an angle of 60°and bisect it. Steps of construction:-
1. Draw a line segment AB.
2. With A as the centre and a small radius, draw an arc cutting AB at P.
3. With M as the centre and the same radius, draw an arc cutting the previously draw arc at Q.
4. Join AQ.
5. With P as the centre and a radius more than half of PQ, draw an arc.
6. With Q as the centre and the same radius as above, draw another arc cutting the previously drawn arc at D.
Thus, AD bisects $\angle$BAC.

#### Question 15:

Construct a ABC in which base BC = 5.2 cm, ∠B = 60° and (AB + AC) = 7.6 cm. Steps of construction:-
1. Draw a line segment BC = 5.2 cm.
2. Construct $\angle$CBX = 60$°$.
3. Set off BP = 7.6 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP at A.
6. Join AC.
Thus, $△$ABC is the required triangle.

#### Question 16:

Construct a square each of whose sides measures 3.2 cm. Steps of construction:-
1. Draw a line segment AB = 3.2 cm.
2. Construct $\angle$BAX = 90$°$ and $\angle$ABY = 90$°$.
3. Set off AD = 3.2 cm and BC = 3.2 cm.
4. Join DC.
Thus, $\square$ABCD is the required square.

#### Question 17:

Construct a ABC in which base BC = 4.8 cm, ∠B = 45° and (ABAC) = 2.5 cm. Steps of construction:-
1. Draw a line segment BC = 4.8 cm.
2. Construct $\angle$CBX = 45$°$.
3. Set off BP = 2.5 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP produced at A.
6. Join AC.
Thus, $△$ABC is the required triangle.

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