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प्रश्न
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
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उत्तर
Let us draw a line segment AM ⊥ BC.
We know that,
Area of a triangle = 1/2 × Base × Altitude
`"Area "(triangleADE)=1/2xxDExxAM`
`"Area "(triangleABD)=1/2xxBDxxAM`
`"Area "(triangleAEC)=1/2xxECxxAM`
It is given that DE = BD = EC
`⇒ 1/2xxDExxAM=1/2xxBDxxAM=1/2xxECxxAM`
⇒ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)
It can be observed that Budhia has divided her field into 3 equal parts.
संबंधित प्रश्न
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is ______.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)
