English

Show That: the Ratio of the Areas of Two Triangles on the Same Base is Equal to the Ratio of Their Heights. - Mathematics

Advertisements
Advertisements

Question

Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

Sum
Advertisements

Solution

Consider the following figure :

Here
Ar. ( ΔABC ) = `1/2` BM x AC
and, Ar. ( ΔADC ) = `1/2` DN x AC

`["Area"( Δ"ABD")]/["Area(Δ ADC )"] = [1/2 "BM" xx "AC"]/[1/2 "DN" xx "AC"]= "BM"/"DN"`

hence proved

shaalaa.com
Figures Between the Same Parallels
  Is there an error in this question or solution?
Chapter 16: Area Theorems [Proof and Use] - Exercise 16 (B) [Page 201]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 16 Area Theorems [Proof and Use]
Exercise 16 (B) | Q 1.3 | Page 201

RELATED QUESTIONS

In the given figure, if the area of triangle ADE is 60 cm2, state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.


The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]


The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB.
Prove that: 
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of the quad. CDEF
= Area of rect. ABDC + Area of // gm. ABEF.


In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)


In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:

(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).


In the following figure, DE is parallel to BC.
Show that: 
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).


In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.


Show that:

A diagonal divides a parallelogram into two triangles of equal area.


In the following figure, BD is parallel to CA, E is mid-point of CA and BD = `1/2`CA
Prove that: ar. ( ΔABC ) = 2 x ar.( ΔDBC )


In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of ΔPOB = 40 cm2, and OP: OC = 1:2, find:
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×