ABCD is a parallelogram, in which AB = 8 cm, AD = 6 cm and altitude AE = 4 cm. Find the altitude corresponding to the side [2 marks]

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Solution

Area of parallelogram ABCD = AB × AE [0.5 mark]
= 8 × 4 cm $^{2}$
= 32 cm $^{2}$ [0.5 mark]

Let altitude corresponding to AD be h.
Then, h × AD = 32 [0.5 mark]
or h × 6 = 32
or h = $\frac{32}{6}$ = $\frac{16}{3}$ [0.5 mark]
Thus, altitude corresponding to AD is $\frac{16}{3}$ cm.

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ABCD is a parallelogram, in which AB = 8 cm, AD = 6 cm and altitude AE = 4 cm. Find the altitude corresponding to the side [2 marks]

Area of parallelogram ABCD = AB × AE [0.5 mark] = 8 × 4 cm2 = 32 cm2 [0.5 mark] Let altitude corresponding to AD be h. Then, h × AD = 32 [0.5 mark] or h × 6 = 32 or h = 326 = 163 [0.5 mark] Thus, altitude corresponding to AD is 163 cm.

Area of parallelogram ABCD = AB × AE [0.5 mark]
= 8 × 4 cm $^{2}$
= 32 cm $^{2}$ [0.5 mark]

Let altitude corresponding to AD be h.
Then, h × AD = 32 [0.5 mark]
or h × 6 = 32
or h = $\frac{32}{6}$ = $\frac{16}{3}$ [0.5 mark]
Thus, altitude corresponding to AD is $\frac{16}{3}$ cm.