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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.2

Question 1.

The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find all the angles.

Solution:

In a quadrilateral the angles add upto 360°.

Let’s call the angles 2x, 4x, 5x, 7x

2x + 4x + 5x + 7x = 360°

18x = 360°

A = 2x = 2 × 20° = 40°

B = 4x = 4 × 20° = 80°

C = 5x = 5 × 20° = 100°

D = 7x = 7 × 20° = 140°

Question 2.

In a quadrilateral ABCD, ∠A = 12° and ∠C is the supplementary of ∠A. The other two angles are 2x – 10 and x + 4. Find the value of x and the measure of all the angles.

Solution:

∠A = 72°

∠C = 180° – 72° (∵ Supplementary at ∠A) = 108°

The other two angles are 2x – 10 and x + 4.

2x – 10 + x + 4 + 108° + 12° = 360°

3x + 174° = 360°

3x = 360° – 174°

∴ ∠A = 72°

∠B = 2x – 10 = 2(62)- 10 = 124 – 10 = 114°

∠C = 108°

∠D = x + 4 = 62 + 4 = 66°

Question 3.

ABCD is a rectangle whose diagonals AC and BD intersect at O. If ∠OAB = 46°, find ∠OBC.

Solution:

∠ABC = 90°

∠OAB + ∠OBC = 90°

46° + ∠OAB = 90°

∠OBC = 90° – 46° = 44°

Question 4.

The lengths of the diagonals of a Rhombus are 12 cm and 16 cm. Find the side of the rhombus.

Solution:

Let ABCD be a rhombus with AC and BD as its diagonals.

We know that the diagonals of a rhombus bisect each other at right angles.

Let O be the intersecting point of both the diagonals

Let AC = 16 cm and BD = 12 cm

use Pythagoras theorem, we have

AB^{2} = OA^{2} + OB^{2}

AB^{2} = 100

∴ AB = 10 cm

Question 5.

Show that the bisectors of angles of a parallelogram form a rectangle.

Solution:

Given ABCD is a parallelogram. Draw the angular bisectors AP, BP, CR and DR of the angles ∠A, ∠B, ∠C and ∠D respectively.

Now to prove : PQRS is a rectangle.

Proof: A rectangle is a parallelogram with one angle 90°.

First we will prove PQRS is a parallelogram.

Now AB || CD and AD is transversal. [ ∴ Interior angles on the same side of transversal are supplementary]

[Opposite sides of a parallelogram are parallel]

Also lines AP and DR intersects

So ∠PSR = ∠DS A

∴ ∠PSR = 90° [∵ Vertically opposite angles]

Similarly we can prove that ∠SPQ = 90°, ∠PQR = 90° and ∠SRQ = 90°

∴ ∠PSR = ∠PQR and ∠SPQ = ∠SRQ

∴ Both pair of opposite angles of PQRS is a parallelogram.

Also ∠PSR = ∠PQR = ∠SPQ = ∠SRQ = 90°

∴ PQRS is a parallelogram with one angle 90°.

∴ PQRS is a rectangle. Hence proved.

Question 6.

If a triangle and a parallelogram lie on the same base and between the same parallels, then prove that the area of the triangle is equal to half of the area of parallelogram.

Solution:

Given: ∆ABE and parallelogram ABCD have the same base and are between the same parallel lines (i.e) l_{1} || l_{2}.

Perpendicular distance between l_{1} and l_{2} = P (say).

Prove that: area of (∆ABE) = \(\frac{1}{2}\) × area of the parallelogram ABCD

Proof: Area of ∆ABE = \(\frac{1}{2}\) × base × height

= \(\frac{1}{2}\) × AB × (Perpendicular distance between l_{1}

= \(\frac{1}{2}\) × AB × P ….(1)

Area of parallelogram ABCD = base × height.

∴ Area of parallelogram ABCD = AB × P …. (2)

From (1) and (2),

Area of ∆ABE = \(\frac{1}{2}\) × Area of parallelogram ABCD.

Hence proved.

Question 7.

Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure. If a || b, c || d, e || f, find the marked angles between

(i) b and c

(ii) d and e

(iii) d and f

(iv) c and f.

Solution:

Since l, m are two parallel lines and PQ, RS, TU, VW are transversal.

Then ∠1 = ∠QOR [vertically opposite angles]

∠1 = 30° [∴ ∠QOR = 30°]

Also, PQ and TU are parallel and m and l are transversal.

Also ∠3 + ∠4 = 180°

⇒ 75° + ∠4 = 180°

∠4 = 180° – 75° = 105°

Hence,

(i) 30°

(ii) 105°

(iii) 75°

(iv) 105°

Question 8.

In the given figure ∠A = 64° , ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ∆ABC, find x° and y°.

Solution:

In ∆ABC, ∠A + ∠B + ∠C = 180°

⇒ 64° + 58° + ∠C = 180°

⇒ 122°+ ∠C = 180°

⇒ ∠C = 180°- 122° = 58°

Also BO and CO are the bisectors of ∠ABC and ∠ACB respectively

Question 9.

In the given Fig. if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ACDF.

Solution:

Given: AB = 2,

BC = 6,

AE = 6,

BF = 8,

CE = 7 and

CF = 7

Consider ∆AEC and ∆BCF,

In ∆AEC,

AC = 8,

AE = 6,

CE = 7

In ∆BCF,

BF = 8,

BC = 6,

CF = 7

∴ ∆AEC ≅ ∆BCF

∴ Area of ∆AEC = Area of ∆BCF

Subtract.area of ∆BDC both sides, we get

Area of ∆AEC – Area of ∆BDC = Area of ∆BCF – Area of ∆BDC

⇒ Area of quadrilateral ABDE = Area of ∆CDF

∴ The required ratio is 1 : 1

Question 10.

In the figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to \(\overline{\mathbf{H E}}\) and \(\overline{\mathbf{F G}}\) ?

Solution:

Area of ABCD = length × breadth.

= DC × BC = 10 × 8 = 80.

Area of ∆AEH = Area of ∆CGF [since they are congruent by RHS rule]

Similarly, Area of ∆BEF = Area of ∆DGH

∴ Area of parallelogram = EFGH = Area of rectangle ABCD – 2(area of ∆AEH) – 2(area of ∆BEF)

Question 11.

In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is \(\frac{9}{8}\) of the area of the parallelogram ABCD.

Solution:

Area of ∆QDP = Area of ∆QMA + Area of ∆MNO + Area of MNBS + Area of ∆MAB

= Area of ∆DCM + Area of ∆MNO + Area of MNBA + Area of ∆NDC

= 2Area of ∆OMN + Area of ∆MNO + 4 Area of ∆OMN + 2 Area of ∆OMN