ABCD is a parallelogram of area 200 cm^2. P is any point on side BC. The two triangles ABP and DPC together with triangle APD fill the parallelogram. Since triangle APD = 1/2 of the parallelogram, the combined area ar(ABP) + ar(DPC) (in cm^2) is:

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100. Triangle APD = 1/2(200) = 100 cm^2 (base AD, apex P on parallel BC). The rest of the parallelogram is ar(ABP) + ar(DPC) = 200 - 100 = 100 cm^2.

ABCD is a parallelogram of area $200 cm^{2}$ . P is any point on side BC. The two triangles ABP and DPC together with triangle APD fill the parallelogram. Since triangle APD = $\frac{1}{2}$ of the parallelogram, the combined area ar(ABP) + ar(DPC) (in $cm^{2}$ ) is:

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More Class 9 Areas — Figures on the Same Base practice

ABCD is a parallelogram of area 200cm2. P is any point on side BC. The two triangles ABP and DPC together with triangle APD fill the parallelogram. Since triangle APD = 21​ of the parallelogram, the combined area ar(ABP) + ar(DPC) (in cm2) is:

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 Areas — Figures on the Same Base practice

ABCD is a parallelogram of area $200 cm^{2}$ . P is any point on side BC. The two triangles ABP and DPC together with triangle APD fill the parallelogram. Since triangle APD = $\frac{1}{2}$ of the parallelogram, the combined area ar(ABP) + ar(DPC) (in $cm^{2}$ ) is: