Model Question Paper-4 Class-7 ML Aggarwal ICSE Mathematics Solutions. APC Understanding Mathematics for ICSE Class-7 Model Question Paper-4 Solutions based on Chapter-10 to 12. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

## Model Question Paper-4 Class-7 ML Aggarwal ICSE Mathematics Solutions

(Based on Chapters 10 to 12)
Time allowed: 1 Hour
Maximum Marks: 25

General Instructions

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

### Paper-4 Class-7 ML Aggarwal

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°

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

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.

In the given figure, two lines l and m intersect each other at O.
∠1 + ∠3 = 222°
⇒ ∠1 = ∠3 (Vertically opposite angles)
⇒ ∠1 = ∠3 = $\frac { 222 }{ 2 }$ = 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.

In the given figure,
AB = AC
∠B = ∠C (Angles opposite to equal sides)
But ∠A = 70°
∠B + ∠C = 180° – 70° = 110°
⇒ ∠B = $\frac { 110 }{ 2 }$
⇒ ∠B = x = 55°

Question 5.
If ∆PQR = ∆EFD, write the parts of ∆EFD that correspond to:
(i) ∠Q
(ii) $\bar { PR }$
(iii) ∠P
(iv) $\bar { QR }$

∆PQR = ∆EFD, then corresponding parts are
(i) ∠Q = ∠F
(ii) $\bar { PR }$ = ED
(iii) ∠P = ∠E
(iv) $\bar { QR }$ = FD

Question 6.
In the given figure, l || m and p || q. Find the value of x, y, and z.

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.

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 = $\frac { 100 }{ 2 }$ = 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?

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.

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?

In the given figure,
In right ∆ABD and ∆ACD
Hypotenuse, AB = AC (Each = 3.6 cm)
∆ABD = ∆ACD (RHS criterion)
BD = DC (c.p.c.t.)
D is mid-point of BC

– : End of Model Question Paper-4 Solved Class-7 ML Aggarwal Solutions  :–