Measuring Space Perimeter and Area Class 9 MCQ Maths Chapter 6

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MCQ on Measuring Space Perimeter and Area Class 9

Measuring Space Perimeter and Area MCQ Class 9

Class 9 Maths Measuring Space Perimeter and Area MCQ

Question 1.
The area of an equilateral triangle with side 2√3 cm is [√3 = 1.732]
(a) 5.196 cm²
(b) 0.866 cm²
(c) 3.496 cm²
(d) 1.732 cm²
Answer:
(a) 5.196 cm²

Explanation:
Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) a²

Question 2.
The area of a rhombus, whose perimeter is 200m and one of the diagonal is 60 m is
(a) 2400 m²
(b) 2100 m²
(c) 2000 m²
(d) 4500 m²
Answer:
(a) 2400 m²

Explanation:
Side, a = \(\frac{200}{4}\) = 50 m
and a² = \(\frac{1}{2}\)
⇒ d₂ = 80 m [given d₁ = 60 m]
Required area = \(\frac{1}{2}\) × d₁ × d₂ = 2400 m²

Question 3.
Each side of an equilateral triangle is 4 cm long. The height of triangle is
(a) 10√3cm
(b) 2√3cm
(c) 4√3 cm
(d) \(\frac{\sqrt{3}}{2}\) cm
Answer:
(b) 2√3cm

Explanation:
Height = \(\frac{\sqrt{3}}{2}\)a²
\(\frac{\sqrt{3}}{2}\)a² = 25√3
⇒ a² = \(\frac{25 \times 4}{\sqrt{3}}\) × √3 = 100
⇒ a = 10 cm

Question 4.
The area of isosceles right angled triangle having each of equal sides a is
(a) \(\frac{1}{2}\)a²
(b) 2a²
(c) a²
(d) \(\frac{\sqrt{2}}{3}\)a²
Answer:
(a) \(\frac{1}{2}\)a²

Explanation:
Area = \(\frac{1}{2}\) × Base × Height = a × a = \(\frac{1}{2}\)a²

Question 5.
If area of equilateral triangle is 25√3 cm² then its perimeter will be
(a) 10 cm
(b) 20 cm
(c) 30 cm
(d) 40 cm
Answer:
(c) 30 cm

Explanation:
Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\)a²
∴ \(\frac{\sqrt{3}}{4}\)a² = 25√3
⇒ a² = \(\frac{25 \times 4}{\sqrt{3}}\) × √3 = 100
⇒ a = 10 cm

Perimeter of equilateral triangle = a + a + a
= 3a = 3 × 10
= 30 cm

Question 6.
The length of hypotenuse of an isosceles right triangle with area 72 cm² is
(a) 12 cm
(b) 12√2cm
(c) 24 cm
(d) 12.5 cm
Answer:
(b) 12√2cm

Explanation:
Area of isosceles right triangle with equal side a
= \(\frac{1}{2}\)a²
Area = \(\frac{1}{2}\) × Base × height
⇒ \(\frac{1}{2}\)a² = 72 × 2
= 144
⇒ a = 12 cm

Hypotenuse of an isosceles right angled triangle
= √2a = 12√2cm

Question 7.
Brahmagupta’s formula is applicable only, when .
(a) all sided are equal
(b) opposite sides are parallel
(c) quadrilateral is cyclic
(d) diagonals are equal
Answer:
(c) quadrilateral is cyclic

Question 8.
The area of a cyclic quadrilateral with sides 5 cm, 6 cm, 7 cm and 8 cm is
(a) \(\sqrt{1680}\)
(b) \(\sqrt{1684}\)
(c) \(\sqrt{144}\)
(d) \(\sqrt{1600}\)
Answer:
(a) \(\sqrt{1680}\)

Question 9.
In Heron’s formula
\(\sqrt{s \times(s-a) \times(s-b) \times(s-c)}\), s is equal to
(a) \(\frac{a+b+c}{4}\)
(b) \(\frac{a+b+c}{3}\)
(c) \(\frac{a+b+c}{2}\)
(d) a + b + c
Answer:
(c) \(\frac{a+b+c}{2}\)

Explanation:
s = \(\frac{a+b+c}{2}\)

Question 10.
The area of triangle with sides 5 cm, 12 cm and 13 cm in length is
(a) 15 cm²
(b) 30 cm²
(c) 60 cm²
(d) 120 cm²
Answer:
(b) 30 cm²

Explanation:
Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
s = \(\frac{5+12+13}{2}\) = 15
Area = \(\sqrt{15 \times 10 \times 3 \times 2}\) = 30 cm²

Question 11.
The sides of a triangle are 25 cm, 17 cm and 12 cm. The length of the altitude on the longest side is equal to
(a) 7.5 cm
(b) 7.2 cm
(c) 8.2 cm
(d) 9.8 cm
Answer:
(b) 7.2 cm

Explanation:
s = \(\frac{1}{2}\) = 27 cm
Area of triangle = \(\frac{1}{2}\)
= \(\frac{1}{2}\) = 90 cm²
Also, area of triangle = \(\frac{1}{2}\) × 25 × h = 90
h = 90 × \(\frac{2}{25}\)
= 7.2 cm

Question 12.
If sides of a triangle are in the ratio 3:5:7 and its perimeter is 150 m. Then, area of triangle is
(a) 100√2 m²
(b) 375√3m²
(c) 750√3m²
(d) 375√2 m²
Answer:
(b) 375√3m²

Explanation:
Sides of a triangle are given by
150 × \(\frac{3}{15}\), 150 × \(\frac{5}{15}\), 150 × \(\frac{7}{15}\) i.e. 30 m, 50 m, 70 m.
Using Heron’s formula,
Area of triangle = 375√3 m²

Question 13.
The difference between the semi-perimeter and the sides of A ABC are 3 cm, 4 cm and 2 cm, respectively. The area of the triangle is
(a) 6√6 cm²
(b) 3√3 cm²
(c) 5√6cm²
(d) 3√6cm²
Answer:
(a) 6√6 cm²

Explanation:
We have, s – a = 3 cm, s – b = 4 cm and s – c = 2 cm
⇒ 3s – (a + b + c) = 9 cm
⇒ 3s – 2s = 9 cm
[∵ \(\frac{a+b+c}{2}\) = s ⇒ a + b + c = 2s]
= 9 cm

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{9 \times 3 \times 4 \times 2}\) = 6√6 cm²

Question 14.
The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm² is
(a) ₹ 2.0
(b) ₹ 2.16
(c) ₹ 2.48
(d) ₹ 3.00
Answer:
(b) ₹ 2.16

Explanation:
Let the edges of the triangular board be a = 6 cm, b = 8 cm and c =10 cm.
Now, semi-perimeter of the triangular board,
s = \(\frac{a+b+c}{2}=\frac{6+8+10}{2}=\frac{24}{2}\) = 12 cm
Now, area of the triangular board
= \(\sqrt{s(s-a)(s-b)(s-c)}\) [by Heron’s formula]
= \(\sqrt{12(12-6)(12-8)(12-10)}\)
= \(\sqrt{12 \times 6 \times 4 \times 2}\)
= \(\sqrt{(12)^2 \times(2)^2}\)
= 12 × 2
= 24 cm²
Since, the cost of painting for area 1 cm² = ₹ 0.09
∴ Cost of paint for area 24 cm² = 0.09 × 24 = ₹ 2.16
Hence, the cost of painting is ₹ 2.16.

Question 15.
The length of the diameter of a circle, whose circumference is 44 cm is
(a) 14 cm
(b) 16 cm
(c) 20 cm
(d) 24 cm
Answer:
(a) 14 cm

Question 16.
The sum of diameters of two circles is 35 cm and the difference of their circumference is 22 cm. The area of the smaller circle will be
(a) 121 cm²
(b) 154 cm²
(c) 309 cm²
(d) 146 cm²
Answer:
(b) 154 cm²

Explanation:
Let the diameters be d₁ and d₂.
d₁ + d₂ = 35
and πd₁ – πd₂ = 22
⇒ π(d₁ – d₂) = 22
⇒ d₁ – d₂ = 7
On solving the Eqs. (i) and (ii), we get d₁ = 21 cm and d₂ = 14 cm
So, radius (r) of smaller circle = 7 cm
Area = πr² = 154 cm²

Question 17.
The area of a circle of radius 14 cm is
(a) 154 cm²
(b) 180 cm²
(c) 616 cm²
(d) 286 cm²
Answer:
(c) 616 cm²

Question 18.
The area of a sector of a circle with radius 7 cm, when the angle of the sector is 60°, is
(a) \(\frac{22}{3}\) cm²
(b) \(\frac{77}{3}\) cm²
(c) \(\frac{44}{3}\) cm²
(d) \(\frac{132}{3}\) cm²
Answer:
(b) \(\frac{77}{3}\) cm²

Explanation:
We have, r =7 cm and angle of sector (0) = 60°
∴ Area of the sector = \(\frac{\theta}{360^{\circ}}\) × πr²
= \(\frac{60^{\circ}}{360^{\circ}}\) × π(7)²
= \(\frac{1}{6} \times \frac{22}{7}\) × 7 × 7
= \(\frac{77}{3}\) cm²

Question 19.
If the area of a sector of a circle is \(\frac{7}{20}\) of the area of the circle then the angle at the centre is equal to
(a) 110°
(b) 130°
(c) 100°
(d) 126°
Answer:
(d) 126°

Explanation:
Let angle at the centre be θ.
We know that
Area of a sector of a circle = \(\frac{\theta}{360^{\circ}}\) × πr²
and area of circle = πr²
According to the question, \(\frac{\theta}{360^{\circ}}\) × πr = \(\frac{7}{20}\) × πr
⇒ \(\frac{\theta}{360^{\circ}}=\frac{7}{20}\)
⇒ θ = \(\frac{7}{20}\) × 360° = 126°

Question 20.
The perimeter of a quadrant of a circle of radius 7 cm, is
(a) 18 cm
(b) 11 cm
(c) 22 cm
(d) 25 cm
Answer:
(d) 25 cm

Explanation:
We have, r = 7 cm

∴ Perimeter of OABCO = 2r + \(\frac{\pi r}{2}\) = 25 cm

Question 21.
OAB is sector of a circle with centre O and radius 7 cm. If length of arc AB = \(\frac {22}{7}\) cm then ∠AOB is equal to
(a) \(\left(\frac{120}{7}\right)^{\circ}\)
(b) 45°
(c) 60°
(d) 30°
Answer:
(c) 60°

Explanation:
Given, radius = 7 cm
and length of arc AB = \(\frac{22}{3}\) cm.
Let ∠AOB = θ
Now, length of arc = \(\frac{1}{2}\) × 2πr
⇒ \(\frac{22}{3}=\frac{\theta}{360^{\circ}}\) × 2 × \(\frac{22}{27}\) × 7
= \(\frac{1}{2}\) = θ
⇒ θ = 60°

Question 22.
The area (in cm²) of a sector of a circle of radius 21 cm cut off by an arc of length 22 cm is
(a) 441
(b) 321
(c) 231
(d) 221
Answer:
(c) 231

Measuring Space Perimeter and Area Class 9 Assertion and Reason Questions

Direction (Q. Nos. 1-7) In the questions given below, there are two statements marked as Assertion (A) and Reason (R). Read the statements and choose the correct option.

Question 1.
Assertion (A): Brahmagupta’s formula gives exact area only for cyclic quadrilaterals.
Reason (R): Opposite angles of cyclic quadrilateral are supplementary.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(b) Both A and R are true but R is not the correct explanation of A.

Explanation:
Assertion Brahmagupta’s formula give exact area only for cyclic quadrilatural and opposite angles of cyclic quadrilateral are supplementary.
Hence, Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 2.
Assertion (A): If the edges of a triangle are 7 cm, 9 cm and 12 cm then the area of triangle is 14√5 cm².
Reason (R): If three sides of a triangle are given then area of triangle cannot always be determined by Heron’s formula.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(c) A is true but R is false.

Explanation:
Assertion Given, sides of a triangle are
Now, s = \(\frac{a+b+c}{2}=\frac{7+9+12}{2}=\frac{28}{2}\) = 14
a = 7 cm, b = 9 cm and c = 12 cm.
∴ Area of triangle using Heron’s formula,
∆ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{14(14-7)(14-9)(14-12)}\)
= \(\sqrt{14 \times 7 \times 5 \times 2}\)
= 14√5 cm²
Hence, Assertion is true.
Reason If three sides of a triangle are given then we can determine the area always by Heron’s formula.
Hence, Reason is false.

Question 3.
Assertion (A): The area of an equilateral triangle with side 2 cm is 3 cm.
Reason (R): If the lengths of the triangle are a, b and c then the area of triangle is \(\sqrt{s(s-a)(s-b)(s-c)}\), where s is the semi-perimeter of the triangle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
Assertion Let the sides of a equilateral triangle be a = 2 cm, b = 2 cm and c = 2 cm
Now, perimeter (s) = \(=\frac{a+b+c}{2}=\frac{2+2+2}{2}\) = 3 cm
Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) [by Heron’s formula]
= \(\sqrt{3(3-2)(3-2)(3-2)}\)
= \(\sqrt{3 \times 1 \times 1 \times 1}\)
= √3 cm²
So, Assertion is false.
Clearly, Reason is true.

Question 4.
Assertion (A): If the outer and inner diameter of a circular path is 10 m and 6 m, respectively then the area of the path isl87tm2.
Reason (R): If R and r be the radius of outer and inner circular path respectively then area of circular path = n(R² – r²).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
Assertion Let R and r be the outer and inner radius, respectively.
R = \(\frac{10}{2}\) = 5 m and r = \(\frac{6}{2}\) = 3 m
Now. area = π(R² – r²)
= (25 – 9)
= 16 π m²
Reason: Correctly explains the formula for area. Hence. Assertion s raise bat Reason is true.

Question 5.
Assertion (A): In the figure, O is the centre of a circle. The area of sector OAPB is \(\frac{5}{18}\) of the area of the circle then the value of x is 100°.

Reason (R): Length of an arc of a circle with radius r and central angle 0 is given by
l = \(\frac{\theta}{360^{\circ}}\) × πr²
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(c) A is true but R is false.

Explanation:
Area of sector OAPB = \(\frac{x}{360^{\circ}}\) × πr²
= \(\frac{5}{18}\)πr²
= \(\frac{x}{360^{\circ}}=\frac{5}{18}\)
⇒ x = 100°
So, Assertion is true.
Now, length of an arc of a circle using radius r and central angle θ,
l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr
So, Reason is false.

Question 6.
Assertion (A): The area of the sector of a circle of radius 5 cm is 9.75 cm2, if the central angle is 90°.
Reason (R): Area of a sector of a circle of radius r and central angle θ is \(\frac{\theta}{360^{\circ}}\) × πr².
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
Let the central angle of the sector be θ.
Given that radius of the sector of circle, r = 5 cm and central angle of the sector, θ = 90°.
Now, area of sector with angle θ = 90°
= \(\frac{90^{\circ}}{360^{\circ}}\) × π × (5)
= \(\frac{1}{4} \times \frac{22}{7}\) × 25
= 19.64 cm²
Hence, Assertion is false but Reason is true.

Question 7.
Assertion (A): If the radius of sector of a circle is reduced to its half and angle is doubled then the perimeter of the sector remains the same.
Reason (R): The length of the arc subtending angle θ at the centre of a circle of radius,
r = \(\frac{\pi r \theta}{180^{\circ}} .\)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
Let r be the radius and θ be the angle of the sector.
Perimeter, P = 2r + rθ
Then radius is reduced to half and angle is ccubled.
Then, perimeter, P’ = 2\(\left(\frac{r}{2}\right)+\frac{r}{2}\) 2θ = r + rθ