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Pythagoras Theorem

In the given ...

Question

In the given figure $s e g D H ⊥ s e g E F a n d ⊥ s e g E F$ . If DH = $12 c m, G K = 20 c m$ and $A (△ D E F) = 300 c m^{2}$ then find (i) EF (ii) $A (△ G E F)$ (iii) $A (□ D F G E)$

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Solution

$(1)$ ar $(△ D E F) = \frac{1}{2} \times E F \times 12 = 300 c m^{2} ⟹ E F = \frac{300 \times 2}{12} = 50 c m$

$(2)$ ar $(△ G E F) = \frac{1}{2} \times E F \times G K = \frac{1}{2} \times 50 \times 30 = 500 c m^{2}$

$(3)$ $a r (□ D F G E) =$ ar $△ D E F$ ar $+ △ G E F = 300 c m^{2} + 500 c m^{2} = 800 c m^{2}$

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Similar questions

s e g D H ⊥ s e g E F

s e g G K ⊥ s e g E F .

D H = 12 c m, G K = 20 c m a n d A (△ D E F) = 300 {c m}^{2}

E F

⊥

⊥

D H = 18 c m

G K = 30 c m

A (Δ D E F) = 300 c m^{2}

E F

A (Δ G E F)

A (□ D E F G)

In fig seg DH ⊥ seg EF and seg GK ⊥ seg EF.
If DH = 12cm, GK= 20 cm
and A(∆DEF) = 300 cm² then find
(i) EF
(ii) A(∆GEF)
(iii) A(⎕DFGE)

In the given figure if $∠ G H E = ∠ D F E = 90^{\circ}$ , DH = 8 cm, DF = 12 cm, DG = 3x - 1 and DE = 4x + 2, then DG and DE are respectively ___.

∠ G H E = ∠ D F E = 90^{\circ}, D H = 8 c m, D F = 12 c m, D G = (3 x - 1) c m

D E = (4 x + 2) c m

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