Question

Find answer of the following figures. Please answer from the given pic!


I’ll mark you as the brainliest.

Answer 1

Question (9) (i) :-

Given :-

• AB & DC = 5 cm = Base of Right angle ∆.
• BD = Length of perpendicular = 7cm.

So,

→ Area of Rt.∆ABD = Area of Rt.∆CDB = (1/2) * Base * Perpendicular

→ Area of Rt.∆ABD = Area of Rt.∆CDB = (1/2) * 7 * 5 = (35/2) cm² .

Hence,

→ Area of Quad.ABCD = Area of Rt.∆ABD + Area of Rt.∆CDB = (35/2) + (35/2) = 2 * (35/2) = 35cm² (Ans.)

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Question (9) (ii) :-

Given :-

• Length of Two parallel sides = 10cm & 14cm.
• Perpendicular Height of Quadrilateral = 7cm.

Solution :-

Since Two opposite sides are parallel to each other , Hence, the given Quadrilateral is a Trapezium.

So ,

→ Area of Trapezium = (1/2) * ( sum of parallel sides) * Perpendicular Height.

→ Required Area = (1/2) * ( 10 + 14) * 7 = (1/2) * 24 * 7 = 12 * 7 = 84cm² . (Ans.)

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Question 10 ) :-

Given :-

• Sides of ∆ are 15, 8 & 17.

Solution :-

Note :- we can solve this problem by 2 methods. As we know that , (8,15,17) are Triplets so its a Right angle ∆.

That will be Easy. Or, we can Solve by Heron's formula.

Lets check first if its a right ∆ or not ?

By Pythagoras Theoram :-

→ Perpendicular² + Base² = Hypotenuse²

→ 15² + 8² = 17²

→ 225 + 64 = 289

→ 289 = 289 ( Hence, Proved).

So, we can say That, Given ∆ is a Right angle ∆.

Hence,

→ Area of Right Angle ∆ = (1/2) * Base * Perpendicular

→ Required Area = (1/2) * 15 * 8 = (1/2) * 120 = 60cm². (Ans.)

________

Now, we also have to Find Length of Altitude Drawn on side 17cm.

Here ,

→ Base will be = 17cm .

→ Altitude = Let x cm.

→ Area = 60cm². (Solved Above).

So,

→ Area of Right Angle ∆ = (1/2) * Base * Altitude

→ (1/2) * x * 17 = 60

→ 17x = 60 * 2

→ 17x = 120

→ x = (120/17) cm. (Ans.)

Hence , Length of Altitude Drawn on side 17cm will be (120/17) cm.

Answer 2

Question 9th ( 1st )

To find :-

Area of these given quadrilaterals .

Given :-

We have two triangles over here ie ABD and DBC . Joining these two we got the given figure .

• Base of triangles = 5cm ie AB as well DC .

• Height of triangles = 7 cm ie DB .

Now it's shown in the Figure that DB makes Right angle on AB and DC both .

→ So these triangles are Right angled triangles .

Area of Right angled triangles = $\frac{1}{2}$ × Base × Height .

→ Area of triangle ABD = $\frac{1}{2}$ × 5 × 7

→ $\frac{35}{2}$ cm²

→ Area of triangle DBC = $\frac{1}{2}$ × 5 × 7 .

→ $\frac{35}{2}$ cm²

Area of ABCD = Area of ABD + Area of DBC .

→ $\frac{35}{2}$ + $\frac{35}{2}$

→ $\frac{35 + 35}{2}$

→ $\frac{70}{2}$

→ Area of ABCD = 35 cm²

Question 9th (2nd )

Given :-

We are provided a trapezium here .

• Parallel sides = 10 cm and 14 cm

• Altitude = 7 cm .

Area of trapezium = $\frac{1}{2}$ × altitude × ( Sum of parallel sides )

→ Area of trapezium = $\frac{1}{2}$ × 7 × (. 10 + 14 )

→ $\frac{1}{2}$ × 7 × ( 24 )

→ 7 × 12 = 84

Area of trapezium = 84 cm²

Question 10th

We have two parts in this question also .

1. Finding area .

2. Finding the length of altitude drawn on side of length 17 cm .

Given :-

• Sides of triangle = 15 cm , 17 cm , 8 cm

Area of triangle = $\sqrt{S ( S- A) . ( S - B) . ( S - C) }$

Here S = Semi perimeter of triangle.

A,B and C are sides of triangle .

→ S = 15 + 17 + 8 / 2

→ S = 40/2

→ S = 20 .

Area = $\sqrt{20(20 - 15) (20 - 17) (20 - 8)}$

→ ( 20 × 5 × 3 × 12 )^ ½

→ ( 3600 ) ^½

→ 60 cm

Area of triangle = 60 cm .

→ Now this triangle is right angled triangle also so we can solve by using that formula too .

Altitude of triangle .

Area of right angle triangle = ½ × base × height.

Now altitude is drawn over side having length 17 cm . So base = 17 cm

→ 60 = ½ × 17 × h

→ 120 = 17 h

→ H = 120/17

So altitude of triangle drawn over side having length of 17 cm is 120/ 17 cm or 7.05 cm