13cm

35°

40°, 60°, 80°

In the figure, two sides of the triangle are equal.
\(\therefore\) Base angles are x and x.
The third angle of the triangle = The vertically opposite angle of 92° = 92°
Now, sum of the three angles of the triangle = 180°
\(\therefore\) x + x + 92° = 180°
or 2x + 92° = 180°
or 2x = 180° - 92° = 88°
or \({2x\over 2}={88°\over2} or x=44°\)
Now, x and y form a linear pair,
\(\therefore\) x + Y = 180°
or 44° + Y = 180°
or y = 180° - 44° =136°.
Thus, x = 44° and y = 136°.

In the figure, two sides of the triangle are equal.
\(\therefore\)The base angles opposite to the equal sides are equal.
Since, one of the base angles is y,
\(\therefore\) Other base angle = y
Now, y and 120° form a linear pair,
\(\therefore\)y + 120° = 180°
or y = 180° - 120° = 60°
Now, sum of the three angles = 180°
\(\therefore\) x + Y + Y = 180°
or x + 60° + 60° = 180°
or x + 120° = 180°
or x = 180° - 120° = 60°
Thus, x = 60° and y = 60°.

Triangle, obtuse angled triangle

According to the question,
Measure of \(\angle B=B\)
Measure of \(\angle A=B-40^0\)
Measure of \(\angle C=B-40^0+50^0\)
We know that, the sum of all three angles in a triangle is equal to 180°.
SO,(B - 40°) + B + (B - 40° + 50°) =180°
\(\Rightarrow\)3B - 30° =180°
\(\Rightarrow\)3B = 210°

\(\therefore B=\frac{210^0}{3}=70^0\)
So, measure of \(\angle A=70^0-40^0=30^0\)

In the given figure, AB = AC, \(\angle A=40°\)
Since, angles opposite to equal sides are also equal.
So, \(\angle B=\angle C\)
We know that, the sum of all angles in a triangle is equal to 180°.
\(\therefore\angle A+\angle B+\angle C=180°\)
\(\Rightarrow40°+2\angle B=180°\)
\(\Rightarrow2\angle B=180°-40°\)
\(\Rightarrow\angle B=\frac{140°}{2}\Rightarrow\angle B=70°\)
\(\therefore\angle B=\angle C=70°\)

Let ABCD be a rectangle, whose length, AB = 40 cm  and diagonal AC = 41 cm
In right angled \(\triangle ABC\), by using Pythagoras property,

AC² = AB² + BC² =>BC² = AC² - AB²
\(\Rightarrow\) BC² = (41)² -(40)²
\(\Rightarrow\) BC² =1681-1600 => BC² =81
\(\Rightarrow BC=\sqrt81=9\)
Now, perimeter of the rectangle = 2 (AB + BC) [∵  perimeter of a rectangle = 2 (l + b)]
= 2 (40 + 9) = 2 \(\times\) 49 =98
Hence, the perimeter of the rectangle is 98 cm.

Let given triangle be \(\triangle ABC\) in which AB = AC. Then, it is an isosceles triangle and given AD is perpendicular to BC, so D is the mid-point of BC.

Given, AB = 37 = AC, AD = 12, BC = x
Now, \(BD=DC=\frac{BC}{2}=\frac{x}{2}\)
​​​​​​In  \(\triangle ADB\) by using Pythagoras properry,
AB² = BD² + AD²
\(\Rightarrow (37)^2=(\frac{x}{2})^2+(12)^2\)
\(\Rightarrow 1369=\frac{x^2}{4}+144\Rightarrow1369-144=\frac{x^2}{4}\)
\(\Rightarrow1225=\frac{x^2}{4}\Rightarrow 1225\times4=x^2\)
\(\Rightarrow\) x² =1225x4 \(\Rightarrow\) x² =4900
\(\Rightarrow x=\sqrt4900=70\)
Hence, the unknown length of x is 70.

\(\angle c=35^0\)

Let the given triangle be \(\triangle ABC\). Then we have AB=BC
So, \(\triangle ABC\) is an isosceles triangle.
\(\therefore\angle A=\angle C\)
[since, the angle opposite to equal sides of an isosceles triangle are equal]

\(\Rightarrow\angle A=x\)
\(\angle B=90^0\)
Now, in \(\triangle ABC\),by angle sum property of a triangle,
\(\angle A+\angle B+\angle C=180^0\)
\(\Rightarrow\) x+90⁰+x=180°
\(\Rightarrow\) 2x+90⁰=180°
\(\Rightarrow\) 2x=180° -90° \(\Rightarrow\) 2x=90°
\(\Rightarrow x=\frac{90^0}{2}=45^0\)
Hence, the value of x is 45°.

\(\angle A+\angle B+\angle C=360^0\)

\(\angle A=90^0\) [Hint Use the concept of is isosceles triangle]

Draw a line segment Be. By paper folding locate the perpendicular bisector of BC.The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and Ae. Thus, the triangle obtained is an isoscelesMBC in which AB = AC.
Since, D is the mid-point of BC, so, AD is its median. Also, AD is perpendicular bisector of Be. So, AD is the altitude of \(\triangle ABC\).
Thus, it is verified that the median and altitude of an isoscelestriangle are same.