# In an isosceles triangle \[ABC\left( {AB = AC} \right)\], the altitude to the base and to the lateral side are equal to 10 cm and 12 cm respectively. The length of the base is

A. 12.5

B. 15

C. 16

D. 18

Answer

Verified

334.8k+ views

**Hint:**In this question, we will be using the formula of half-angles in trigonometry and using the sides of the triangle we will obtain the unknown sides. Also, we will use the property of altitude to the base in case of isosceles triangle to get the required solution. So, use this concept to reach the solution of the given problem.

**Complete step by step answer:**

Given \[\Delta ABC\] is an isosceles triangle with \[AB = AC\]. Let \[AD\] be the altitude to the base and \[CE\] be the altitude to the lateral side whose length are equal to 10 cm and 12 cm respectively as shown in the below figure:

By applying \[\sin B\] in \[\Delta ABD\] and \[\Delta BCE\], we get

\[

\Rightarrow \sin B = \dfrac{{AD}}{{AB}} = \dfrac{{CE}}{{BC}} \\

\Rightarrow \sin B = \dfrac{{10}}{{AB}} = \dfrac{{12}}{{BC}} \\

\Rightarrow \dfrac{{10}}{{AB}} = \dfrac{{12}}{{BC}} \\

\Rightarrow BC = \dfrac{{12}}{{10}}AB \\

\therefore BC = 1.2AB \\

\]

We know that the altitude to the base divides the angle at where the two equal side lengths meet into two equal angles.

So, \[\angle BAC = \dfrac{1}{2}\angle BAD = \dfrac{1}{2}\angle DAC\]

We know that altitude to the base in an isosceles triangle divides the base in two equal parts.

So, we have \[BC = 2BD = 2DC\]

Now in \[\Delta ABD\] we have

\[

\Rightarrow \sin \dfrac{A}{2} = \dfrac{{BD}}{{AB}} = \dfrac{{\dfrac{{BC}}{2}}}{{AB}} = \dfrac{{\dfrac{{1.2AB}}{2}}}{{AB}} = 0.6{\text{ }}\left[ {\because BC = 1.2AB} \right] \\

\Rightarrow \cos \dfrac{A}{2}{\text{ = }}\sqrt {1 - {{\left( {0.6} \right)}^2}} = 0.8{\text{ }}\left[ {\because \cos \dfrac{A}{2} = \sqrt {1 - {{\sin }^2}\dfrac{A}{2}} } \right] \\

\Rightarrow \tan \dfrac{A}{2}{\text{ = }}\dfrac{{BD}}{{AD}} = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}} = \dfrac{{0.6}}{{0.8}} = \dfrac{3}{4} = 0.75 \\

\therefore \dfrac{{BD}}{{AD}} = 0.75 \Rightarrow BD = 10 \times 0.75 = 7.5{\text{ }}\left[ {\because AD = 10} \right] \\

\]

We know that altitude to the base in an isosceles triangle divides the base in two equal parts.

So, \[BC = 2 \times BD = 2 \times 7.5 = 15\]

Therefore, the length of the base is \[BC = 15{\text{cm}}\]

**So, the correct answer is “Option B”.**

**Note:**We have to use a trigonometric formula to solve this question. The formula used is \[\cos \dfrac{A}{2} = \sqrt {1 - {{\sin }^2}\dfrac{A}{2}} \]. Also, we have to use \[\sin A = \dfrac{{{\text{opp side}}}}{{{\text{hypotenuse}}}}\] and \[\tan \dfrac{A}{2} = \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}\]. Also, we should remember the property of the isosceles triangle that opposite sides are always equal and opposite angles are also equal.

Last updated date: 23rd Sep 2023

•

Total views: 334.8k

•

Views today: 9.34k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is the IUPAC name of CH3CH CH COOH A 2Butenoic class 11 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

The dimensions of potential gradient are A MLT 3A 1 class 11 physics CBSE

Define electric potential and write down its dimen class 9 physics CBSE

Why is the electric field perpendicular to the equipotential class 12 physics CBSE