In the above triangle ABC, point P is the centroid of the triangle. The length of AE is 15 cm. Find the length of AP and PE.

AP = 10 cm, PE = 5 cm

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AP = 9 cm, PE = 6 cm

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AP = 8 cm, PE = 7 cm

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AP = 5 cm, PE = 10 cm

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Solution

The correct option is A
AP = 10 cm, PE = 5 cm

The point of intersection of all the medians (or the centroid), divides each median in the ratio 2 : 1.
So, AP : PE =2 : 1
New, AE =15
$A P = \frac{2}{3} \times 15 = 10 c m$
$P E = \frac{1}{3} \times 15 = 5 c m$

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In the above triangle ABC, point P is the centroid of the triangle. The length of AE is 15 cm. Find the length of AP and PE.

The correct option is A AP = 10 cm, PE = 5 cm The point of intersection of all the medians (or the centroid), divides each median in the ratio 2 : 1. So, AP : PE =2 : 1 New, AE =15 AP=23×15=10cm PE=13×15=5cm

The correct option is A
AP = 10 cm, PE = 5 cm

The point of intersection of all the medians (or the centroid), divides each median in the ratio 2 : 1.
So, AP : PE =2 : 1
New, AE =15
$A P = \frac{2}{3} \times 15 = 10 c m$
$P E = \frac{1}{3} \times 15 = 5 c m$