Respuesta :

GIven: A Quadrilateral ABCD in which [tex]\frac{Area\triangle ABD}{Area\triangle ACD} =\frac{1}{2}[/tex].BH is an altitude in ΔABD and  CL is an altitude in △ACD.

To Find: BH : CL

Solution: We know, Area of a triangle = 1/2 × Base × Altitude

Area (Δ A B D)= 1/2 × BH × AD, where BH is altitude and AD is base.-----(1)

Area (Δ A CD) = 1/2 × CL × AD  ,where CL is an altitude and AD is base.

                                                --------------------------------(2)

Dividing (1) by (2), we get

→[tex]\frac{Area\triangle ABD}{Area\triangle ACD} = \frac{BH}{CL}[/tex]

As, [tex]\frac{Area\triangle ABD}{Area\triangle ACD}=\frac{1}{2}[/tex]

So, [tex]\frac{BH}{CL}= \frac{1}{2}[/tex]

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Answer:

BH:CL=1:2

Step-by-step explanation:

We are given that a quadrilateral ABCD,

[tex]\frac{Ar\triangle ABD}{Ar\triangle ACD}=\frac{1}{2}[/tex]

BH is an altitude in triangle ABD and CL is an altitude in triangle ACD.

We have to find the BH:CL

We know that

Area of triangle=[tex]\frac{1}{2}\times base\times height[/tex]

Area of triangle ABD=[tex]\frac{1}{2}\times AD\times BH[/tex]

Area of triangle ACD=[tex]\frac{1}{2}\times AD\times CL[/tex]

Substitute the values then we get

[tex]\frac{\frac{1}{2}\times AD\times BH}{\frac{1}{2}\times AD\times CL}=\frac{1}{2}[/tex]

[tex]\frac{BH}{CL}=\frac{1}{2}[/tex]

Hence, BH:CL=1:2

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