Advertisements
Advertisements
प्रश्न
Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the base is 14 cm.
उत्तर
We know that the altitude drawn from the vertex opposite to the non-equal side bisects the non-equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
`AB^2=BD^2+DA^2`
⇒ `25^2=7^2+DA^2`
⇒ `DA^2=625-49`
⇒ `DA^2=576`
⇒ `DA=24 cm`
APPEARS IN
संबंधित प्रश्न
In the following figure, DE || AC and DF || AE. Prove that `("BF")/("FE") = ("BE")/("EC")`
In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC
If AD = 8x − 7, DB = 5x − 3, AE = 4x − 3 and EC = (3x − 1), find the value of x.
In the given figure, side BC of a ΔABC is bisected at D
and O is any point on AD. BO and CO produced meet
AC and AB at E and F respectively, and AD is
produced to X so that D is the midpoint of OX.
Prove that AO : AX = AF : AB and show that EF║BC.
In the given figure, O is a point inside a ΔPQR such that ∠PQR such that ∠POR = 90°, OP = 6cm and OR = 8cm. If PQ = 24cm and QR = 26cm, prove that ΔPQR is right-angled.
ΔABC is am equilateral triangle of side 2a units. Find each of its altitudes.
Find the height of an equilateral triangle of side 12cm.
Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.
State and converse of Thale’s theorem.
◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.
In fig, seg DE || sec BC, identify the correct statement.