Ratio of areas of two triangles with equal height is `2:3`. If base of the smaller triangle is 6 cm, the what is the corresponding base of the bigger triangle?

Ratio of areas of two triangles with equal height is `2:3`. If base of

1.	Ratio of areas of two triangles with equal height is `2:3`. If base of the smaller triangle is 6 cm, the what is the corresponding base of the bigger triangle?
Answer» Let the areas of the two triangles be `A_(1)` and `A_(2)`. Let their respective bases be `b_(1)` and `b_(2)`. `A_(1):A_(2)=2:3` and `b_(1)=6cm`…(Given) The triangles are of equal heights. Areas of triangles with equal heights are proportional to their corresponding bases. `:.(A_(1))/(A_(2))=(b_(1))/(b_(2)):.2/3=6/(b_(2)):.b_(2)=(6xx3)/2` `:.b_(2)=9cm` The corresponding base of the bigger triangle is 9cm.

Discussion

No Comment Found

Related InterviewSolutions

Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of `/_QPR`.
In the figure, `PS=3`, `SQ=6`, `QR=5`, `PT=x` and `TR=y`. Give any two pairs of values of `x` and `y` such that line `ST||` side `QR`.
As shown in the adjoining figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m, then how long will be the shadow of the bigger pole at the same time?
`DeltaLMN~DeltaPQR`, `9xxA(DeltaPQR)=16xxA(DeltaLMN)`. If `QR=20`, then find `MN`.
As shown in the figures, two poles of height 8m and 4m are perpendicular to the groudn. If the length of shadow smaller pole due to sunlight is 6 m then long will be the shadow of the bigger pole of the same time?
In the adjoining figure, `BP bot AC`, `CQ bot AB`, `A-P-C`, `A-Q-B`. Complete the following activity to Prove `DeltaAPB~DeltaAQC`. In `DeltaAPB` and `DeltaAQC` `/_APB=square^(@)`…….`(1)` `/_AQC=square^(@)`……..`(2)` `:./_APB~=/_AQC`…....(`square`) `:.DeltaAPB~DeltaAQC`......(`square`)
In `DeltaMNP, NQ` is a bisector of `/_N`. If `MN=5, PN=7, MQ=2.5,` the find QP.
In the figure seg `PQ||` seg DE, `A(DeltaPQF)=20` units `PF=2DP`,then find `A(square DPQE)` by completing the following activity: Activity: `A(DeltaPQF)=20` sq units, `PF=2DP`. Let us assume `DP=x` `:.PF=2x` `DF=DE+square=square+square=3x` In `DeltaFDE` and `DeltaFPQ`. `/_FDE~=/_square` ..........(Corresponding angles) `/_FED~=/_square` .....(Corresponding angles) `:.DeltaFDE~DeltaFPQ` .....(AA test) `:.(A(DeltaFDE))/(A(DeltaFPQ))=(square)/(square)=((3x)^(2))/((2x)^(2))=9/4` `A(DeltaFDE)=9/4A(DeltaFPQ)=9/4xxsquare=square` `A(squareDPQE)=A(DeltaFDE)-A(DeltaFPQ)` `=square-square` `=square`
In figure `XY||` seg AC. If `2AX=3BX` and `XY=9`, complete the activity to find the value of AC. Activity: `2AX=3BX` `:.(AX)/(BX)=(square)/(square)` `:.(AX+BX)/(BX)=(square+square)/(square)`.........(By componendo) `:.(AB)/(BX)=(square)/(square)`.............1`DeltaBCA~DeltaBYX`.......(`square` test of similarity) `:.(BA)/(BX)=(AC)/(XY)` ..............(Corresponding sides of similar triangles) `:.(square)/(square)=(AC)/9` `:.AC=square` ...........[From 1]
In the adjoining figure, seg `XY||` seg `AC`, IF `3AX=2BX` and `XY=9` then find the length of `AC`.

Ratio of areas of two triangles with equal height is `2:3`. If base of the smaller triangle is 6 cm, the what is the corresponding base of the bigger triangle?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment