The two triangles are similar.
1 answer:
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Answer:
133
Step-by-step explanation:
To solve the problem we could separate the figure into three parts. First figure is a triangle, second figure is a rectangle, third figure is a triangle. See image attached.
Solve each area of the figures
First figure, a triangle that have 7 units long of the base, and 2 units long of the height.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the first figure is 7 units²
Second figure is a rectangle, the length of the rectangle is 7 units, the width of the rectangle is 4 units.
a = l × w
a = 7 × 4
a = 28
The area of the second figure is 28 units²
Third figure is a triangle, the base is 7 units long and the height is 2 units long.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the third figure is 7 units²
The area of the three figures
area = first figure area + second figure area + third figure area
area = 7 + 28 + 7
area = 42
The total area of the figures is 42 units²
Solve each area of the figures
First figure, a triangle that have 7 units long of the base, and 2 units long of the height.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the first figure is 7 units²
Second figure is a rectangle, the length of the rectangle is 7 units, the width of the rectangle is 4 units.
a = l × w
a = 7 × 4
a = 28
The area of the second figure is 28 units²
Third figure is a triangle, the base is 7 units long and the height is 2 units long.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the third figure is 7 units²
The area of the three figures
area = first figure area + second figure area + third figure area
area = 7 + 28 + 7
area = 42
The total area of the figures is 42 units²
Answer:
The following measurements are:
(Option #4)
(Option #7)
(Option #5)
(Option #2)
Step-by-step explanation:
To begin, we can find the measure of by applying the inscribed angle theorem: an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle.
Since the intercepted arc (RS) is 46 degrees, we have:
Next, we can find the measure of arc QT using the same theorem. So,
Notice that the chord RT is actually a diameter. From the theorem about the inscribed angle including a diameter, we know that the intercepted arc will have a measure of . Since the arc ST is part of the arc RST, and we know RS is
, we can set up and solve this equation:
We can use the same idea to find RQ. We know that RQT is and QT is
, so: