Equal-area theorems, base-altitude reasoning MCQs

Q1. ABCD is a parallelogram. P is any point on diagonal BD. Triangles APD and CPD are compared. Which statement is true?

• A.ar(APD) = ar(CPD)
• B.ar(APD) = 2·ar(CPD)
• C.ar(APD) = (1/2)·ar(CPD)
• D.They are unequal in general

Answer: A. ar(APD) = ar(CPD)

NCERT fact (math, chapter 'Areas of Parallelograms and Triangles'): BD is a diagonal of parallelogram ABCD, so ar(ABD) = ar(CBD). Subtracting ar(BPD) from both equal areas (since P lies on BD, BPD is common to triangles ABD and CBD when viewed correctly), or directly: triangles APD and CPD share base PD and have equal heights from A and C to line BD because diagonals bisect the parallelogram. Hence ar(APD) = ar(CPD). Final answer - A) ar(APD) = ar(CPD).

Q2. Triangle ABC has area 90. Points D, E lie on BC with BD = DE = EC. Lines AD and AE divide the triangle into three smaller triangles. Which statement is true?

• A.All three have equal area 30
• B.Areas are 20, 30, 40
• C.Areas are 15, 30, 45
• D.Areas are 22.5, 45, 22.5

Answer: A. All three have equal area 30

Q3. In a triangle ABC, the medians AD, BE, CF intersect at G. ar(AGF) equals what fraction of ar(ABC)?

• A.1/6
• B.1/3
• C.1/4
• D.1/12

Answer: A. 1/6

Start by listing the data - the numerical data stated in the question (math, chapter 'Areas of Parallelograms and Triangles'). What we must find: the requested quantity. The principle that connects these is - So ar(AGF) = (1/6)·ar(ABC). Substituting and simplifying: So ar(AGF) = (1/6)·ar(ABC). The three medians of a triangle divide it into 6 smaller triangles of equal area. That lands on option A) 1/6.