## 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°

Question 2.

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

(a) scalene

(b) isosceles

(c) equilateral

(d) right-angled

Question 3.

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

Question 4.

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

Question 5.

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

(i) ∠Q

(ii)

(iii) ∠P

(iv)

Question 6.

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

Question 7.

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

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?

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.

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?

Solutions

Solution 1:

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)

Solution 2:

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)

Solution 3:

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°

Solution 4:

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°

Solution 5:

∆PQR = ∆EFD, then corresponding parts are

(i) ∠Q = ∠F

(ii) = ED

(iii) ∠P = ∠E

(iv) = FD

Solution 6:

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°

Solution 7:

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°

Solution 8:

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

Solution 9:

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.

Solution 10:

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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