Advertisements
Advertisements
प्रश्न
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
Advertisements
उत्तर
Consider ΔACD.
Line-segment CD is bisected by AB at O. Therefore, AO is the median of
ΔACD.
∴ Area (ΔACO) = Area (ΔADO) ... (1)
Considering ΔBCD, BO is the median.
∴ Area (ΔBCO) = Area (ΔBDO) ... (2)
Adding equations (1) and (2), we obtain
Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO)
⇒ Area (ΔABC) = Area (ΔABD)
APPEARS IN
संबंधित प्रश्न
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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)
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)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
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.]
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
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 ______.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
