## ML Aggarwal Class 7 ICSE Maths Model Question Paper 4

(Based on Chapters 10 to 12)

**Time allowed: 1 Hour**

**Maximum Marks: 25**

**General Instructions**

- Questions 1-2 carry 1 mark each.
- Questions 3-5 carry 2 marks each.
- Questions 6-8 cany 3 marks each
- Questions 9-10 carry 4 marks each.

Choose the correct answer from the given four options (1-2):

Question 1.

In the given figure, if l || m then the value of x is

(a) 65°

(b) 105°

(c) 115°

(d) 125°

Solution:

In the given figure, if l || m

∠1 = 65° (Vertically opposite angles)

But ∠1 + x = 180° (Co-interior angles)

⇒ 65° + x = 180°

⇒ x = 180° – 65° = 115° (c)

Question 2.

A triangle whose two angles measure 40° and 100° is

(a) scalene

(b) isosceles

(c) equilateral

(d) right-angled

Solution:

In a triangle, two angles are 40° and 100°

Third angle = 180° – (40° + 100°) = 180° – 140° = 40°

The triangles is an isosceles as two angles are equal,

then their opposite sides are equal. (b)

Question 3.

In the given figure, lines l and m intersect at O. If ∠1 + ∠3 = 222°, then find the measure of ∠2.

Solution:

In the given figure, two lines l and m intersect each other at O.

∠1 + ∠3 = 222°

⇒ ∠1 = ∠3 (Vertically opposite angles)

⇒ ∠1 = ∠3 = = 111°

But ∠1 + ∠2 = 180° (Linear pair)

⇒ 111° + ∠2 = 180°

⇒ ∠2 = 180° – 111°

⇒ ∠2 = 69°

Question 4.

In the given figure, AB = AC. Find the value of x.

Solution:

In the given figure,

AB = AC

∠B = ∠C (Angles opposite to equal sides)

But ∠A = 70°

∠B + ∠C = 180° – 70° = 110°

⇒ ∠B =

⇒ ∠B = x = 55°

Question 5.

If ∆PQR = ∆EFD, write the parts of ∆EFD that correspond to:

(i) ∠Q

(ii)

(iii) ∠P

(iv)

Solution:

∆PQR = ∆EFD, then corresponding parts are

(i) ∠Q = ∠F

(ii) = ED

(iii) ∠P = ∠E

(iv) = FD

Question 6.

In the given figure, l || m and p || q. Find the value of x, y, and z.

Solution:

In the given figure,

l || m and p || q

x = 70° (Corresponding angles)

∠1 = x = 75° (Corresponding angles)

But ∠1 + y = 180° (Co-interior angles)

y = 180° – ∠1 = 180° – 75° = 105°

z = ∠1 (Vertically opposite angles)

z = 75°

Hence, x = 75°, y = 105°, z = 75°

Question 7.

In the given figure, AB = AC. Find the values of x and y.

Solution:

In the given figure,

AB = AC

∠EAF = 80°

∠BAC = ∠EAF (Vertically opposite angles)

∠1 = 80°

∠B + ∠C = 180° – ∠A = 180° – 80° = 100°

But ∠B = ∠C (Angles opposite to equal sides)

∠B = ∠C = = 50°

x = 50°

Now Ext. ∠ACD = ∠B + ∠BAC = x + ∠1

y = 50° + 80° = 130°

x = 50°, y = 130°

Question 8.

If the lengths of two sides of a triangle are 5 cm and 7.5 cm then what can be the length of the third?

Solution:

Lenghts of two sides of a triangle are 5 cm, 7.5 cm

Then length of third side < (5 + 7.5) = 12.5 cm

or greater than 7.5 – 5.0 = 2.5 cm

Question 9.

An apple orchard is in the shape of a rectangle. If its length is 60 m and the length of one diagonal is 75 m, then find:

(i) the breadth of the orchard.

(ii) the perimeter of the orchard.

Solution:

An apple orchard is in shape of a rectangle Length = 60 cm

One diagonal = 75 m

and perimeter = 2(Length + Breadth)

= 2(60 + 45) m = 2 × 105 = 210 m

Apples keep our body hale and healthy.

Question 10.

In the given figure, measures of some parts are given.

(i) State the three pairs of equal parts in ∆ABD and ∆ACD.

(ii) Is ∆ABD = ∆ACD? Give reason.

(iii) Is D mid-point of BC? Why?

Solution:

In the given figure,

In right ∆ABD and ∆ACD

Side AD = AD (Common)

Hypotenuse, AB = AC (Each = 3.6 cm)

∆ABD = ∆ACD (RHS criterion)

BD = DC (c.p.c.t.)

D is mid-point of BC.

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