Target Exam

CUET

Subject

General Aptitude Test

Chapter

Quantitative Reasoning

Topic

Mensuration: 2D/3D

Question:

Match the given areas of various solid objects with their respective formulae:

List I Area of solid object

List II Formula

(A) Lateral surface area of a cylinder

(I) $\pi r\sqrt{r^2+h^2}$

(B) Total surface area of a hemisphere

(II) $4\pi r^2$

(C) Lateral surface area of a cone

(III) $2\pi rh$

(D) Surface area of a sphere

(IV) $3πr^2$

(Where, r = radius of the solid shape (or base) and h = height of the solid shape.)

Choose the correct answer from the options given below:

Options:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation:

The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

List I Area of solid object

List II Formula

(A) Lateral surface area of a cylinder

(III) $2\pi rh$

(B) Total surface area of a hemisphere

(IV) $3πr^2$

(C) Lateral surface area of a cone

(I) $\pi r\sqrt{r^2+h^2}$

(D) Surface area of a sphere

(II) $4\pi r^2$

(A) Lateral Surface Area of a Cylinder: The lateral (curved) surface of a cylinder is found by multiplying the circumference of the base ($2\pi r$) by the height ($h$). Formula: $2\pi rh$ → Matches with (III)

(B) Total Surface Area of a Hemisphere: A hemisphere has a curved surface area ($2\pi r^2$) and a flat circular base ($\pi r^2$). The total surface area is the sum of both: $2\pi r^2 + \pi r^2 = 3\pi r^2$. Formula: $3\pi r^2$ → Matches with (IV)

(C) Lateral Surface Area of a Cone: The lateral surface area is given by $\pi r l$, where $l$ is the slant height. The slant height can be calculated using the Pythagorean theorem as $l = \sqrt{r^2 + h^2}$. Formula: $\pi r \sqrt{r^2 + h^2}$ → Matches with (I)

(D) Surface Area of a Sphere: The total surface area of a sphere is exactly four times the area of its circular cross-section. Formula: $4\pi r^2$ → Matches